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Existence Theorems in Linear Chaos | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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Existence Theorems in Linear Chaos

Stanislav Shkarin*

Queens's University Belfast, Pure Mathematics Research Centre, University Road, Belfast, UK

Corresponding Author:
Stanislav Shkarin
Queens's University Belfast
Pure Mathematics Research Centre
University Road, Belfast, BT7 1NN, UK
Tel: +442890245133
E-mail: [email protected]

Received date: July 21, 2015; Accepted date: August 03, 2015; Published date: August 31, 2015

Citation: Shkarin S (2015) Existence Theorems in Linear Chaos. J Generalized Lie Theory Appl S1:009. doi:10.4172/generalized-theory-applications.S1-009

Copyright: © 2015 Shkarin S. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

Chaotic linear dynamics deals primarily with various topological ergodic properties of semigroups of continuous linear operators acting on a topological vector space. In this survey paper, we treat questions of characterizing which of the spaces from a given class support a semigroup of prescribed shape satisfying a given topological ergodic property.
In particular, we characterize countable inductive limits of separable Banach spaces that admit a hypercyclic operator, show that there is a non-mixing hypercyclic operator on a separable infinite dimensional complex Fréchet space X if and only if X is non-isomorphic to the space ω of all sequences with coordinatewise convergence topology. It is also shown for any k ∈ N, any separable infinite dimensional Fréchet space X non-isomorphic to ω admits a mixing uniformly continuous group {Tt}t∈Cn T of continuous linear operators and that there is no supercyclic strongly continuous operator semigroup {Tt}t≥0 on ω. We specify a wide class of Fréchet spaces X, including all infinite dimensional Banach spaces with separable dual, such that there is a hypercyclic operator T on X for which the dual operator T′ is also hypercyclic. An extension of the Salas theorem on hypercyclicity of a perturbation of the identity by adding a backward weighted shift is presented and its various applications are outlined.

Keywords

Hypercyclic operators; Mixing semigroups; Backward weighted shifts; Bilateral weighted shifts

Introduction

Unless stated otherwise, all vector spaces in this article are over the field equation, being either the field equation of complex numbers or the fieldequation of real numbers, all topological spaces are assumed to be Hausdorff and all vector spaces are assumed to be non-trivial. As usual, equation equation is the set of integers,equation is the set of non-negative integers,equation is the set of positive integers and equation is the set of non-negative real numbers. Symbol L(X, Y) stands for the space of continuous linear operators from a topological vector space X to a topological vector space Y. We write L(X) instead of L(X, X) and X′ instead of L(X, equation). For each T ∈ L(X), the dual operator T′: X′ → X′ is defined as usual: (T′ f)(x)=f (Tx) for f ∈ X′ and x ∈ X. By a quotient of a topological vector space X we mean the space X/Y, where Y is a closed linear subspace of X. We start by recalling some definitions and facts.

Notation and definitions

A topological vector space is called locally convex if it has a base of neighborhoods of zero consisting of convex sets. Equivalently, a topological vector space is locally convex if its topology can be defined by a family of seminorms. For brevity, we say locally convex space for a locally convex topological vector space. A subset B of a topological vector space X is called bounded if for any neighborhood U of zero in X, a scalar multiple of U contains B. We say that τ is a locally convex topology on a vector space X if (X, τ) is a locally convex space. If X is a vector space and Y is a linear space of linear functionals on X separating points of X, then the weakest topology on X, with respect to which all functionals from Y are continuous, is denoted σ(X, Y). The elements of X can be naturally interpreted as linear functionals on Y, which allows one to consider the topology σ(Y, X) as well. If equation is a family of bounded subsets of (Y, σ(Y, X)), whose union is Y, then the seminorms

equation

define the topology on X of uniform convergence on sets of the family equation. The topology on X of uniform convergence on all bounded subsets of (Y, σ(Y, X)) is called the strong topology and denoted β(X, Y). The topology of uniform convergence on all compact convex subsets of (Y, σ(Y, X)) is called the Mackey topology and is denoted τ (X, Y). According to the Mackey-Arens theorem, for a locally convex space (X, τ) and a space Y of linear functionals on X, the equality Y=X′ holds if and only if σ(X, Y) ⊆ τ ⊆ τ (X, Y). We say that a locally convex space (X, τ) carries a weak topology if τ coincides with σ(X, Y) for some space Y of linear functionals on X, separating points of X. If X is a locally convex space, we write Xβ for (X′, β(X, X′)), Xτ for (X, τ (X, X′)) and Xσ for (X, σ(X, X′)). Similarly we denote (X, β(X′, X)) by Xβ′, (X′, τ (X′, X)) by Xτ′ and (X′, σ(X′, X)) by Xσ′. An equation-space is a complete metrizable topological vector space. A locally convex equation-space is called a Fréchet space. If {Xα: α ∈ A} is a family of locally convex spaces, then their (locally convex) direct sum is the algebraic direct sum equation of the vector spaces Xα endowed with the strongest locally convex topology, which induces the original topology on each Xα. Let equation be a sequence of vector spaces such that Xn is a subspace of Xn+1 for each n ∈ Z+ and each Xn carries its own locally convex topology τn such that τn is (maybe nonstrictly) stronger than the topology τn+1|Xn. Then the inductive limit of the sequence {Xn} is the space equation endowed with the strongest locally convex topology τ such that τ|Xn ⊆ τn for each n ∈equation. In other words, a convex set U is a neighborhood of zero in X if and only if U ∩ Xn is a neighborhood of zero in Xn for each n ∈equation. An LB-space is an inductive limit of a sequence of Banach spaces. An LBs-space is an inductive limit of a sequence of separable Banach spaces. We use symbol φφ to denote the locally convex direct sum of countably many copies of the one- dimensional space equation and the symbol ω to denote the product of countably many copies of equation. Note that φ is a space of countable algebraic dimension and carries the strongest locally convex topology (=any seminorm on φ is continuous). We can naturally interpret ω as the space equation of all sequences with coordinatewise convergence topology. Clearly ω is a separable Fréchet space. Recall also that if X is a locally convex space and A ⊂ X′, then A is called uniformly equicontinuous if there exists a neighborhood U of zero in X such that |f (x)| ≤ 1 for any x ∈ U and f ∈ A.

Let T be a continuous linear operator on a topological vector space X. A vector x ∈ X is called a cyclic vector for T if the linear span of the orbit O(T, x)={Tn: n ∈ equation} of x is dense in X. The operator T is called cyclic if T has a cyclic vector. Recall also that for n ∈ equation, T is called n-cyclic if there are vectors x1,...,xn ∈ X such that the linear span of the set {Tnxj : n ∈ equation, 1 ≤ j ≤ n} is dense in X. Obviously, 1-cyclicity coincides with cyclicity. We say that T is multicyclic if it is n-cyclic for some n ∈ equation.

Let X and Y be topological spaces and {Ta: a ∈ A} be a family of continuous maps from X to Y. An element x ∈ X is called universal for this family if the orbit {Tax: a ∈ A} is dense in Y and {Ta: a ∈ A} is said to be universal if it has a universal element. We say that a family {Tn : n ∈ equation} is hereditarily universal if any its infinite subfamily is universal. An operator semigroup on a topological vector space X is a family {Tt}t∈A of elements of L(X) labeled by elements of an abelian monoid A (monoid is a semigroup with identity) and satisfying T0=I, Ts+t=TtTs for any t, s ∈ A (unless stated otherwise, we use additive notation for the operation on A). A norm on A is a function | • | : A → [0, ∞) satisfying |na|=n|a| and |a + b| ≤ |a| + |b| for any for any n ∈ equation and a, b ∈ A. An abelian monoid equipped with a norm will be called a normed semigroup. We will be mainly concerned with the case when A is a closed (additive) subsemigroup of equation containing 0 with the norm |a| being the Euclidean distance from a to 0. In the latter case we consider A to be equipped with topology inherited from equation and we say that an operator semigroup {Tt}t∈A is strongly continuous if the map equation from A to X is continuous for any x ∈ X. We say that an operator semigroup {Tt}t∈A is uniformly continuous if there exists a neighborhood U of zero in X such that for any sequence equation of elements of A converging to t∈A, Ttnx converges to Ttx uniformly on U. Clearly, uniform continuity is strictly stronger than strong continuity. It is also worth mentioning that many authors use the term ’uniformly continuous semigroup’ for semigroups satisfying the weaker of uniform convergence of Ttnx to Ttx on any bounded subset of X.

If A is a normed semigroup and {Tt}t∈A is an operator semigroup on a topological vector space X, then we say that {Tt}t∈A is (topologically) transitive if for any non-empty open subsets U, V of X, the set {|t| : t ∈ A, Tt(U)∩V ≠∅} is unbounded. We say that {Tt}t∈A is (topologically) mixing if for any non-empty open subsets U, V of X, there is r=r(U,V) > 0 such that Tt(U)∩V≠∅ provided |t| > r. We also say that {Tt}t∈A is hypercyclic (respectively, supercyclic) if the family {Tt:t ∈ A} (respectively, {zTt: z ∈ equation, t ∈ A}) is universal. {Tt}t∈A is said to be hereditarily hypercyclic (respectively, hereditarily supercyclic) if for any sequence equation of elements of A such that |tn| → ∞, the family {Ttn:n ∈ equation} (respectively, {zTtn:z ∈ equation, n ∈ equation}) is universal. A continuous linear operator T acting on a topological vector space X is called hypercyclic, supercyclic, hereditarily hypercyclic, hereditarily supercyclic, mixing or transitive if the semigroup equation has the same property. It is worth noting that our definition of a hereditarily hypercyclic operator follows Ansari [1], while in the terminology of references [2,3], the same property is called ’hereditarily hypercyclic with respect to the sequence nk=k of all nonnegative integers’. Hyper- cyclic and supercyclic operators have been intensely studied during last few decades, [4-6] and references therein. Clearly mixing implies transitivity and hereditary hypercyclicity (respectively, hereditary supercyclicity) implies hypercyclicity (respectively, supercyclicity). Recall that a topological space X is called a Baire space if the intersection of countably many dense open subsets of X is dense in X. According to the classical Baire theorem, complete metric spaces are Baire.

Proposition 1.1. Let X be a topological vector space, A be a normed semigroup and S={Ta}a∈A be an operator semigroup on X. Then

(1.1.1) if equation is hereditarily hypercyclic, then equation is mixing.

If additionally X is Baire separable and metrizable, the converse implication holds:

(1.1.2) if equation is mixing, then equation is hereditarily hypercyclic.

The above proposition is a combination of well-known facts, appearing in the literature in various modifications. It is worth noting that a similar statement holds for hypercyclicity and transitivity under certain natural additional assumptions. One can also write down and prove a supercyclicity analogue of the above proposition. In the next section we shall prove Proposition 1.1 for sake of completeness. It is worth noting that for any subsemigroup A0 of A, not lying in the kernel of the norm, {Tt}t∈A0 is mixing if {Tt}t∈A is mixing. In particular, if {Tt}t∈A is mixing, then Tt is mixing whenever |t| > 0.

Results

The question of existence of supercyclic or hypercyclic operators or semigroups on various types of topological vector spaces was intensely studied. There are no hypercyclic operators on any finite dimensional topological vector space and there are no supercyclic operators on a finite dimensional topological vector space of real dimension > 2. These facts follow, for instance from the main result of reference [7]. Herzog [8] demonstrated that there is a supercyclic operator on any separable infinite dimensional Banach space. Later Ansari [9] and Bernal-Gonzáles [10], answering a question raised by Herrero, showed independently that for any separable infinite dimensional Banach space X there is a hypercyclic operator T ∈ L(X). Using the same idea as in reference [9], Bonet and Peris [11] proved that there is a hypercyclic operator on any separable infinite dimensional Fréchet space and demonstrated that there is a hypercyclic operator on an inductive limit X of a sequence Xn for n ∈ equation of separable Banach spaces provided there is n ∈ equation for which Xn is dense in X. Grivaux [3] observed that hypercyclic operators T constructed in references [9-11] are in fact mixing and therefore hereditarily hypercyclic. They actually come from the same source. Namely, according to Salas [12] an operator of the shape I + T, where T is a backward weighted shift on equation, is hypercyclic. Virtually the same proof as in reference [12] demonstrates that these operators are in fact mixing. Moreover, all operators constructed in the above papers, except for the ones acting on ω, are hypercyclic because of a quasisimilarity with one of the operators of the shape identity plus a backward weighted shift. The same quasisimilarity transfers the mixing property as effectively as it transfers hypercyclicity. A similar idea was used by Bermúdez, Bonilla and Martinón [13] and Bernal- González and Grosse-Erdmann [14], who have demonstrated that any separable infinite dimensional Banach space supports a hypercyclic strongly continuous semigroup {Tt}t∈R+. Bermúdez, Bonilla, Conejero and Peris [15] have shown for any separable infinite dimensional complex Banach space X, there exists a mixing strongly continuous semigroup {Tt}t∈Π with Π={z ∈ equation : Re z ≥ 0} such that the map equation is holomorphic on the interior of Π. As a matter of fact, one can easily see that the semigroup constructed in reference [15] extends to a holomorphic mixing group equation. Finally, Conejero [16] proved that any separable infinite dimensional complex Fréchet space nonisomorphic to ω supports a uniformly continuous mixing operator semigroup {Tt}t∈R+.

The following theorem extracts the maximum of the method both in terms of the class of spaces and semigroups. Although the general idea remains the same, the proof requires dealing with a number of technical details of various nature. In particular, we will prove and apply a multi- operator version of the Salas theorem. For brevity we shall introduce the following class of locally convex spaces.

Definition 1.2. We say that a sequence equation of elements of a topological vector space X is an equation-sequence if xn → 0 in X and the series equation converges in X for each a ∈ equation.

We say that a locally convex space X belongs to the class M if its topology is not weak and there exists an equation-sequence in X with dense span.

Theorem 1.3. Let X ∈ M. Then for any k ∈ equation, there exists a hereditarily hypercyclic (and therefore mixing) uniformly continuous operator group equation. Moreover, if equation=equation the map equationfrom equationk to X is holomorphic for each x ∈ X.

Since for any hereditarily hypercyclic semigroup equation and any non-zero t ∈equationk, the operator Tt is hereditarily hypercyclic, we have the following corollary.

Corollary 1.4. Let X ∈ M. Then there is a hereditarily hypercyclic (and therefore mixing) operator T ∈ L(X).

Remark 1.5. It is easy to see that if X ∈ M, then Xτ ∈ M. Indeed, if equation is an equation-sequence in X with dense span, then equation is an equation-sequence in Xτ with dense span. Of course, Xσ never belongs to M. On the other hand, it is well-known that L(Xσ)=L(Xτ). Moreover, since σ(X, X′) ⊆ τ(X, X′), then any strongly continuous hereditarily hypercyclic operator semigroup equation on Xτ is also strongly continuous and hereditarily hypercyclic as an operator semigroup on Xσ. Thus in the case Xτ ∈ M, Theorem 1.3 implies that there is a strongly continuous hereditarily hypercyclic operator semigroup equation on Xσ. Unfortunately, the nature of the weak topology does not allow to make such a semigroup uniformly continuous.

It is worth noting that any separable Fréchet space admits an equation- sequence with dense span. It is also well-known [17] that any Fréchet space carries the Mackey topology and the topology on a Fréchet space X differs from the weak topology if and only if X is infinite dimensional and is non-isomorphic to ω. That is, any separable infinite dimensional Fréchet space non-isomorphic to ω belongs to M. Similarly, one can verify that an infinite dimensional inductive limit X of a sequence Xn for n ∈ equation of separable Banach spaces belongs to M provided there is n ∈ equation for which Xn is dense in X. Thus all the above mentioned existence theorems are particular cases of Theorem 1.3.

Grivaux [3] raised a question whether each separable infinite dimensional Banach space supports a hypercyclic non-mixing operator. Since the class M contains separable infinite dimensional Banach spaces, the following theorem provides an affirmative answer to this question.

Theorem 1.6. Let X ∈ M. Then there exists T ∈ L(X) such that T is hypercyclic and non-mixing.

The simplest separable infinite dimensional locally convex space space (and the only Fréchet space) outside M is ω. Curiously, the situation with ω is totally different. Hypercyclic operators on the complex space ω have been characterized by Herzog and Lemmert [18]. Namely, they proved that a continuous linear operator T on the complex Fréchet space ω is hypercyclic if and only if the point spectrum σp(T′) of T′ is empty. It also worth mentioning that Bés and Conejero [19] provided sufficient conditions for T ∈ L(ω) to have an infinite dimensional closed linear subspace, each non-zero vector of which is hypercyclic, and found common hypercyclic vectors for some families of hypercyclic operators on ω. See also the related work [20] by Petersson. The following theorem extends the result of Herzog and Lemmert and highlights the difference between ω and other Fréchet spaces.

Theorem 1.7. Let T ∈ L(ω) be such that T′ has no non-trivial finite dimensional invariant subspaces and equation be a sequence of polynomials such that deg pl → ∞ as l → ∞. Then the family {pl (T) : l ∈ equation} is universal. Moreover, there is no strongly continuous supercyclic semigroup equation on ω.

Note that in the case equation=equation, T′ has no non-trivial finite dimensional invariant subspaces if and only if σp(T′)=Φ. The first part of the above theorem implies that any hypercyclic operator on ω is mixing. We shall, in fact, verify the following more general statement.

Theorem 1.8. Let X be a locally convex space carrying weak topology and T ∈ L(X). Then the following conditions are equivalent

(1.8.1) T′ has no non-trivial finite dimensional invariant subspaces;

(1.8.2) T is transitive;

(1.8.3) T is mixing;

(1.8.4) the semigroup equation is mixing, where equation is the multiplicative semigroup of non-zero polynomials with the norm |p|=deg p.

Remark 1.9. Chan and Sanders [21] observed that on the space (equation)σ, being the Hilbert space equation with the weak topology, there is a transitive non-hypercyclic operator. Theorem 1.8 provides a huge supply of such operators. For instance, the backward shift T on equation is mixing on (equation)σ (T′ has no non-trivial finite dimensional invariant subspaces) and T is clearly non-hypercyclic (each its orbit is bounded).

Theorems 1.3, 1.6 and 1.7 imply the following curious corollary.

Corollary 1.10. Let X be a separable infinite dimensional Fréchet space. Then the following are equivalent

(1.10.1) there is a hypercyclic non-mixing operator T ∈ L(X);

(1.10.2) there is a mixing uniformly continuous semigroup equation on X ;

(1.10.3) there is a supercyclic strongly continuous semigroup equation on X;

(1.10.4) X is non-isomorphic to ω.

Another simple space outside M is φ. Bonet and Peris [11] observed that there are no supercyclic operators on φ. On the other hand, Bonet, Frerick, Peris and Wengenroth [22] constructed a hypercyclic operator on the locally convex direct sum equation of countably many copies of the Banach space equation. The space X is clearly an LBs-space, is complete and non-metrizable. It is also easy to see that equation (there are no equation-sequences in X with dense span). We find sufficient conditions of existence and of non-existence of a hypercyclic operator on a locally convex space. These conditions allow us to characterize the LBs-spaces, which admit a hypercyclic operator.

Theorem 1.11. Let X be the inductive limit of a sequence equation of separable Banach spaces. Then the following conditions are equivalent:

(1.11.1) X admits no hypercyclic operator;

(1.11.2) X admits no cyclic operator with dense range;

(1.11.3) X is isomorphic to Y ×φ, where Y is the inductive limit of a sequence equation of separable

Banach spaces such that Y0 is dense in Y;

(1.11.4) for any sufficiently large n, equation is finite dimensional and the set equation is infinite, where equation is the closure of Xk in X.

The proof is based upon the following result, which is of independent interest.

Theorem 1.12. Let X be a topological vector space, which has no quotients isomorphic to φ. Then there is no cyclic operator with dense range on X ×φ.

The following theorem provides another generalization of the mentioned result of Bonet, Frerick, Peris and Wengenroth.

Theorem 1.13. Let equation be a sequence of separable Fréchet spaces. Then there is a hypercyclic operator on equation if and only if the set {n ∈ equation : Xn is infinite dimensional} is infinite.

We derive the above theorem from the following result, concerning more general spaces.

Theorem 1.14. Let Xn ∈ M for each n ∈ equation and equation Then there is a hypercyclic operator on X.

The next issue, we discuss, are dual hypercyclic operators. Let X be a locally convex space. Recall that Xβ′ is the dual space X′ endowed with the strong topology β(X′, X). It is worth noting that if X is a normed space, then the strong topology on X′ coincides with the standard norm topology. Salas [23] has constructed an example of a hypercyclic operator T on equation such that both T and T′ are hypercyclic. This result motivated Petersson [24] to introduce the following definition. We say that a continuous linear operator T on a locally convex space X is dual hypercyclic if both T and T′ are hypercyclic on X and Xβ′ respectively. Using the construction of Salas, Petersson proved that any infinite dimensional Banach space X with a monotonic and symmetric Schauder basis and with separable dual admits a dual hypercyclic operator. He also raised the following questions. Does there exist a dual hypercyclic operator on any infinite dimensional Banach space with separable dual? Does there exist a non-normable Fréchet space that admits a dual hypercyclic operator? The first of these questions was recently answered affirmatively by Salas [25]. The following theorem provides a sufficient condition for existence of a dual hypercyclic operator on a locally convex space.

Theorem 1.15. Let X be an infinite dimensional locally convex space admitting an equation-sequence with dense span. Assume also that there is an equation-sequence equation with dense span in Xβ′ and at least one of the following conditions is satisfied:

(1.15.1) the topology of X coincides with σ(X, X′);

(1.15.2) the topology of X coincides with τ (X, X′);

(1.15.3) the set {fn : n ∈ equation} is uniformly equicontinuous.

Then X admits a dual hypercyclic operator.

Since every separable Fréchet space admits an equation-sequence with dense span and every Fréchet space carries the Mackey topology, the above theorem implies the following corollary.

Corollary 1.16. Let X be a separable infinite dimensional Fréchet space, such that there is an equation-sequence with dense span in Xβ′. Then there exists a dual hypercyclic operator T ∈ L(X).

If X is a Banach space, Xβ′ is also a Banach space and therefore has an equation-sequence with dense span if and only if it is separable. Thus Corollary 1.16 implies the next corollary, which is the mentioned recent result of Salas.

Corollary 1.17. Let X be an infinite dimensional Banach space with separable dual. Then there exists a dual hypercyclic operator T ∈ L(X).

Corollary 1.16 also provides plenty of non-normable Fréchet spaces admitting a dual hypercyclic operator, thus answering the second of the above questions of Petersson. For instance, take the complex Fr´echet space X of entire functions on one variable with the topology of uniform convergence on compact sets. It is easy to verify that the sequence of functionals gn(f)=(n!)−1f(n)(0) is an equation-sequence with dense span in Xβ′. Since X is also infinite dimensional and separable,Corollary 1.16 implies that X supports a dual hypercyclic operator.

The proofs of the above results are based upon the two main ingredients. One of them are sufficient conditions of mixing and the other is a criterion for a generic (in the Baire category sense) operator from a given class to be hypercyclic. Our sufficient conditions of mixing extend the result of Salas on hypercyclicity of perturbations of the identity by adding a backward weighted shift. Apart from providing us with tools, these extensions are of independent interest.

Theorem 1.18. Let X be a topological vector space and T ∈ L(X) be such that the space

equation (1.1)

 

is dense in X. Then T is mixing. If additionally, X is Baire, separable and metrizable, then T is hereditarily hypercyclic.

We shall see that the above theorem implies not only the mentioned result of Salas, but also is applicable in many other situations. For instance, we use the above theorem to prove the following results.

Theorem 1.19. Let X be a separable infinite dimensional Banach space and equation be the operator norm closure in L(X) of the set of finite rank nilpotent operators. Then the set of T ∈ equation for which T is supercyclic and I + T is hypercyclic is a dense Gδ subset of the complete metric space equation. If additionally X′ is separable, then the set of T ∈ equation for which T and T′ are supercyclic and I + T and I + T′are hypercyclic is a dense Gδ subset of equation.

Note that if a Banach space X has the approximation property [26], then the set equation from the above corollary is exactly the set of compact quasinilpotent operators (in the case equation=equation by quasinilpotency of T we mean quasinilpotency of the complexification of T or equivalently that ||Tn||1/n → 0). Thus we have the following corollary.

Corollary 1.20. Let X be a separable infinite dimensional Banach space with the approximation property and equation ⊂ L(X) be the set of compact quasinilpotent operators. Then the set of T ∈ equation for which T is supercyclic and I + T is hypercyclic is a dense Gδ subset of the complete metric space equation. If additionally X′ is separable, then the set of T ∈ equation for which T and T′ are supercyclic and I + T and I + T′ are hypercyclic is a dense Gδ subset of equation.

Theorem 1.21. Let X be a separable infinite dimensional Banach space and equation be the set of nuclear quasinilpotent operators endowed with the nuclear norm metric. Then the set of T ∈ equation for which T is supercyclic and I + T is hypercyclic is a dense Gδ subset of the complete metric space equation. If additionally X′ is separable, then the set of T ∈ equation for which T and T′ are supercyclic and I + T and I + T′ are hypercyclic is a dense Gδ subset of equation.

Theorems 1.21 and 1.19 provide a large supply of dual hypercyclic operators T on any infinite dimensional Banach space with separable dual.

Extended Backward Shifts

Godefroy and Shapiro [27] have introduced the notion of a generalized backward shift. Namely, a continuous linear operator T on a topological vector space X is called a generalized backward shift if its generalized kernel

equation

is dense in X and ker T is one-dimensional. We introduce a more general concept. Namely, we say that T is an extended backward shift if

equation (2.1)

 

is dense in X. From the easy dimension argument [27] it follows that if T ∈ L(X) is a generalized backward shift, then dimker Tn=n and T (kerTn+1)=ker Tn for each n ∈ equation. Hence ker Tn=Tn (ker T2n) and therefore ker Tn=Tn (X) ∩ ker Tn for any n ∈ equation. It follows that ker*T=kerT for a generalized backward shift. That is, any generalized backward shift is an extended backward shift.

We also consider the following analog of the concept of an extended backward shift for a k-tuple of operators. Let T1,...,Tk be continuous linear operators on a topological vector space X. We say that T=(T1,...,Tk) ∈ L(X)k is a EBSk-tuple if TmTj=TjTm for any j,m ∈ {1,..., k} and is dense in X.

equation (2.2)

 

It is easy to see that in the case of one operator (that is, k=1 and T=T1 ∈ L(X)), κ (n,T) = Tn (kerT2n)=kerTn ∩Tn(X) and therefore the last definition is a generalization of the previous one. In order to study extended backward shifts we need to establish some properties of the backward shift on the finite dimensional space equation2n.

Backward shift on equation2n

The following lemma is a modification of a lemma from reference [28].

Lemma 2.1. For each n ∈ equation and z ∈ equation\{0}, the matrix equation is invertible.

Proof. For each n, k ∈ equation consider the matrix equation First, we demonstrate that the determinants of Mn,k satisfy the recurrent formula

equation (2.3)

 

The equality (2.3) for n=2 is trivial. Suppose now that n ≥ 3. Subtracting the previous column from each column of Mn,k except the first one, we see that det Mn,k=det Nn,k where equation. Dividing the j-th row of Nn,k by j and multiplying the j-th column by aj=(k + n − j + 1)(k + n − j) for 1 ≤ j ≤ n, we arrive at the matrix Mn−1,k+2. Hence

equation

which proves (2.3). Since det M1,k=1 for each k ∈ equation, from (2.3), it follows that det Mn,k ≠ 0 for any n, k ∈ equation. Let now Bn be the matrix obtained from An,1 by putting the columns of An,1 in the reverse order. Clearly det An,1=(−1)n−1 det Bn. On the other hand, multiplying the j-th column of Bn by (n − j + 1)! for 1 ≤ j ≤ n, we get the matrix Mn,1. Hence,

equation

Since det Mn,1 ≠ 0, we see that det An,1 ≠ 0 and therefore An,1 is invertible. Finally, for any z ∈ equation consider the diagonal n × n matrix Dn,z with the entries (1, z,..., zn−1) on the main diagonal. It is straightforward to verify that

An,z=zDn,z An,1Dn,z for any z ∈ equation. (2.4)

Since An,1 and Dn,z for z ≠ 0 are invertible, we see that An,z is invertible for each n ∈ equation and z ∈equation\ {0}.

Lemma 2.2. Let n ∈ equation and e1,...,e2n be the canonical basis of equation2n and S ∈ L(equation2n) be the backward shift defined by Se1=0 and Sek=ek−1 for 2 ≤ k ≤ 2n and P the linear projection on equation2n onto the subspace E=span {e1,...,en} along F=span {en+1,...,e2n}. Then for any z ∈equation\ {0} and u, v ∈ E, there exists a unique xz=xz(u, v) ∈ equation2n such that

Pxz=u and PezSxz=v. (2.5)

Moreover, for any bounded subset B of E and any ε > 0, there is c=c(ε, B) > 0 such that

equation (2.6)

 

equation (2.7)

 

In particular, xz (u, v) → u and ezS xz(u, v) → v as |z| → ∞ uniformly for u, v from any bounded subset of E.

Proof. Let u, v ∈ E and z ∈ equation\ {0}. For y ∈ equation2n we denote

equation

One easily sees that (2.5) is equivalent to the vector equation

equation (2.8)

 

where equation is the matrix from Lemma 2.1 and equation is defined as

equation

provided we set xj=uj for 1 ≤ j ≤ n. According to Lemma 2.1, the matrix equation is invertible for any equation and therefore (2.8) is uniquely solvable. Thus there exists a unique equation satisfying (2.5). It remains to verify the estimates (2.6) and (2.7). From (2.9) it follows that for any bounded subset B of E and any ε > 0, there is a=a(ε, B) such that

equation (2.10)

 

Recall that Dn,z is the diagonal n × n matrix with the entries (1, z,...,zn−1) on the diagonal. Equalities (2.8) and (2.4) imply

equation

where we use invertibility of An,1 provided by Lemma 2.1. According to (2.10), the set equation is bounded in equation Hence the set equation is bounded in equation. From the last display we see that

equation

Boundedness of Q implies now that (2.6) is satisfied with some c=c1(ε, B). Finally, since

equation

there exists c=c2(ε, B) for which (2.7) is satisfied. Hence both (2.7) and (2.6) are satisfied with

c=max{ c1(ε, B), c2(ε, B)}.

The next corollary follows immediately from Lemma 2.2.

Corollary 2.3. Let equation and equation be as in Lemma 2.2. Then for any u, v ∈ E and any sequence equation satisfying equation there exists a sequence equation of elements of equation such that xj → u and equation

Lemma 2.4. Let equation and equation be as in Lemma 2.2. Then for any bounded sequences equation of elements of E, there exists a sequence equation of elements of equation such that xj − uj → 0 and

Proof. It is easy to see that there is equation such that J has an upper triangular matrix, J is invertible and equation Indeed, S and eS − I are similar since they are nilpotent of maximal rank 2n − 1. Moreover, since S and eS − I are upper triangular, the similarity operator can be chosen upper triangular and therefore J (E) ⊆ E.

Let now xj=J−1 xj (Juj, Jvj), where xz (u, v) is defined in Lemma 2.2. Since the set equation

is bounded and is contained in E because J (E) ⊆ E, from Lemma 2.2 it follows that xj (Juj, Jvj) − Juj → 0 and ejS xj (Juj, Jvj) − Jvj → 0 as j → ∞. Multiplying by J−1, we obtain xj − uj → 0 and J−1 ejS Jxj(I+S)j xj − vj → 0 as j → ∞.

In order to construct multi-parameter mixing semigroups, we need the following multi-operator version of Corollary 2.3.

Lemma 2.5. Let equation for each j ∈ {1,..., k} let equation be the canonical basis in equation equation be the backward shift: Sj ej1=0 and 2 ≤ l ≤ 2nj. Let also equation and for 1 ≤ j ≤ k,

equation

where Sj sits in the jth place. Finally, let equation be a sequence of elements of equation with equation Then for any u, v ∈ E, there exists a sequence equation of elements of X such that xm → u and

equation

Proof. If the statement of the lemma is false, then there are u, v ∈ E and a subsequence equation such that (u, v) does not belong to the closure of the set equation

Let equation be the one-point compactification of equation. Since equation is compact and metrizable, we can pick a convergent in equation subsequence equation Clearly the statement of the lemma remains false with {zm} replaced by {zm”}. That is, it suffices to consider the case when {zm} converges in equation.

Thus without loss of generality, we can assume that {zm} converges to w ∈ equation . Let C={j : wj=∞}. Since equation the set C is non-empty. Without loss of generality, we may also assume that C={1,..., r} with 1 ≤ r ≤ k.

Denote by Σ the set of (u, v) ∈ X × X such that there exists a sequence equation of elements of X for which xm → u and equation We have to demonstrate that E × E ⊆ Σ. Let uj ∈ Ej for 1 ≤ j ≤ k and equation By Corollary 2.3, for 1 j ≤ r, there exist sequences equation of elements of equation such that

equation

Now we put equation and equation Consider the sequences equation and equation of elements of X defined by the formula equation According to the definition of xm and ym and the above display, xm → 0 and ym → u. Indeed, xm → 0 because for any j, the sequence xj,m is bounded and x1,m → 0. Similarly, taking into account that (zm)j → wj for j > r, we see that equation . Hence (u, 0) ∈ Σ and (0, u) ∈ Σ. Thus

equation

where equation On the other hand, it is easy to see that the linear span of the set ({0} × E0) ∪ (E0 × {0}) is exactly E × E. Since Σ is a linear space, the above display implies that E × E ⊆ Σ.

For applications it is more convenient to reformulate the above lemma in the coordinate form.

Corollary 2.6. Let equation for each j ∈ {1,..., k}let Nj={1,..., 2nj}, Qj ={1,..., nj}. Consider the sets M=N1 ×... × Nk, M0=Q1 ×... × Qk and let {em : m ∈ M} be the canonical basis of the finite dimensional vector space equation Let E=span {em : m ∈ M0} and for 1 ≤ j ≤ k, Tj ∈ L(X) be the operator acting on the canonical basis in the following way: Tjem=0 if mj=1, Tj em=em′ if mj > 1, where ml′=m′l if l ≠ j, mj′=mj − 1. Then for any sequence equation of elements of equation and any u, v ∈ E, there exists a sequence equation of elements of X such that xm → u and equation

The key lemmas

Lemma 2.7. Let X be a Hausdorff topological vector space, equation and A=(A1,..., Ak) ∈ L(X)k be such that Aj Al=Al Aj for any l, j ∈ {1,..., k}. Then for each x from the space κ(n, A) defined by (2.2), there exists a common finite dimensional invariant subspace for A1,..., Ak such that for any sequence equation of elements of equation there exist sequences equation, equation of elements of Y for which

equation (2.11)

 

where equation

Proof. Since x ∈ κ(n, T), there exists y ∈ X such that equation y = 0 for 1 ≤ j ≤ k. For each j ∈ {1,..., k} let Nj={1,..., 2nj} and Qj={1,..., nj}. Denote M=N1 ×... × Nk, M0=Q1 ×... × Qk. For any l ∈ M, let equation yand let Y=span {hl : l∈ M}. Clearly Y is finite dimensional. It is also straightforward to verify that Ajhl=0 if lj=1 and Aj hl=hl′ if lj > 1, where l′r=lr for r ≠ j, l′j=lj − 1, 1 ≤ j ≤ k0. It follows that Y is invariant for Aj for 1 ≤ j ≤ k. Consider the linear operator equation defined on the canonical basis by the formulas Jel=hl for l ∈ M. Let also E=span {el : l ∈ M0} and equation be the operators from Corollary 2.6. Taking into account the definition of Tj and the action of Aj on hl, we see that AjJ=JTj for 1 ≤ j ≤ k. Clearly n=(n1,..., nk) ∈ M0 and therefore en ∈ E. Since equation , we also see that x=hn. According to Corollary 2.6, there exist sequences equation of elements of equation such that equation and equation . Now equation for equation Then {xm} and {ym} are sequences of elements of Y. From the intertwining relations AjJ=JTj and the fact that equation and Y are finite dimensional, it follows that equationequation, equation Thus (2.11) is satisfied.

Let X be a topological vector space and equation We write T ∈ ε(k, X) if for ∞ any equation the series equation converges in X and the map equation from equation is separately continuous, where equation and equation . The following proposition is an elementary exercise. We leave its proof for the reader.

Proposition 2.8. equation X be a locally convex space and T ∈ ε (k, X). Assume also that Tj Tm = Tm Tj for any m, j ∈ {1,..., k}. Then equation is a strongly continuous operator group. Moreover, if equation then the map equation is holomorphic for each x ∈ X.

Remark 2.9. It is worth noting that the semigroup property equation fails if the operators Tj are not pairwise commuting.

Corollary 2.10. Let X be a locally convex space, equation and A=(A1,..., Ak) ∈ ε (k, X) be such that Aj Al=Al Aj for any l, j ∈ {1,..., k}. Then for each x, y from the space ker (A) defined by (2.2) and any sequence equation of elements of equation there exist a sequence equation in X such that um → x and equation as m → ∞.

Proof. Fix a sequence equation of elements of equation Let Σ be the set of (x, y) ∈ X × X for which there exists a sequence equation in X such that um → x and equation as m → ∞. According to Lemma 2.7, κ(n, A) × {0} ⊆ Σ and {0} × κ(n, A) ⊆ Σ for any n=(n1,..., nk) ∈ equation where the space κ(n, A) is defined by (2.2). On the other hand, from the definition of Σ it is clear that Σ is a linear subspace of X × X. Thus

equation

Hence (x, y) ∈ Σ for any x, y ∈ ker† (A).

Lemma 2.11. Let X be a topological vector space,
equation There exist sequences equation of elements of X such that

equation (2.12)

 

Proof. If x=0, we can take uk=vk=0, so we may assume that x ≠ 0. Let n be the smallest positive integer for which Anx=0. Since Amx=0, we have n ≤ m. Hence x ∈ Am(X) ⊆ An(X). Thus we can pick w ∈ X such that Anw=x. Denote

equation

Clearly Ahj=hj-1 for 2 ≤ j ≤ 2n, hn=Anh2n=Anw=x and Ah1=A2nh2n=Anx=0. By definition of n we have h1=A2n−1h2n=An−1x ≠ 0. In particular equation Since the order of nilpotency of a nilpotent operator on Y cannot exceed the dimension of Y, we have dim Y ≥ 2n. On the other hand Y is the span of the 2n-elements set {h1,..., h2n}. Hence {h1,..., h2n} is a linear basis of Y. Thus there exists a unique linear isomorphism equation such that Jek=hk for 1 ≤ j ≤ 2n. Since Ahj=hj−1 for 2 ≤ j ≤ 2n and Ah1=0, we have A|y=JSJ−1, where equation is the backward shift operator from Lemma 2.2. Applying Lemma 2.4 with um=(0,..., 0, 1) and vm=(0,..., 0, 0), we find that there exists a sequence equation of vectors in equation such that gk → en and (I + S)kgk → 0 as k → ∞. Applying Lemma 2.4 with um=(0,..., 0, 0) and vm=(0,..., 0, z−m), there exists also a sequence equation of vectors in equation such that fk → 0 and zk (I + S)kfk → en as k → ∞. Define now uk=Jfk and vk=Jgk for equation Since Y and equation are finite dimensional, we see that

equation

Thus the sequences {uk} and {vk} satisfy the desired conditions.

The following corollary of Lemma 2.11 seems to be of independent interest.

Theorem 2.12. Let X be a topological vector space, T ∈ L(X) and Λ(T) be the set defined in (1.1). Then for any u, v ∈ Λ(T), there exists a sequence equation of elements of X such that xk → x and Tk xk → y. Pick equation
equation

By Lemma 2.11, there exist sequences equation in X satisfying (2.12). Since I + A=z−1 T, (2.12) can be rewritten in the following way:

equation

This shows that for any equation
equation

On the other hand, the fact that Σ is a linear subspace of X × X, the definition of Λ(T) and the above display imply that Λ(T) × Λ(T) ⊆ Σ.

Mixing semigroups and extended backward shifts

We start by proving Proposition 1.1. The next observation is Proposition 1 in reference [28].

Proposition G. Let X be a topological space and equation be a family of continuous maps from X to X such that TαTβ=TβTα for any α, β ∈ A and Tα(X) is dense in X for any α ∈ A. Then the set of universal elements for F is either empty or dense in X.

The following general theorem can be found in reference [28].

Theorem U. Let X be a Baire topological space, Y a second countable topological space and {Ta: a ∈ A} a family of continuous maps from X into Y. Then the following assertions are equivalent:

(U1) The set of universal elements for {Ta : a ∈ A} is dense in X;

(U2) The set of universal elements for {Ta : a ∈ A} is a dense Gδ -subset of X;

(U3) The set {(x, Tax): x ∈ X, a ∈ A} is dense in X × Y.

Proof of Proposition 1.1. Assume that {Tt}t∈A is hereditarily hypercyclic. That is, equation is universal for any sequence equation of elements of A such that |tn| → ∞. Applying this property to the sequence tn=nt with t ∈ A, |t| > 0, we see that Tt is hypercyclic. Since any hypercyclic operator has dense range [28], we get that Tt(X) is dense in X for any t ∈ A with |t| > 0. We proceed by reasoning ad absurdum. Assume that {Tt}tA is non-mixing. Then there exist nonempty open subsets U and V of X and a sequence equation of elements of A such that |tn| → ∞ and equation for each equation Since each Ttn has dense range and Ttn commute with each other, Proposition G implies that the set W of universal elements of equation is either empty or dense in X. Since equation is universal, W is non-empty and therefore dense in X. Hence we can pick x ∈ W ∩ U. Since x is universal for equation , there is equation for which Ttn x ∈ V. Hence equation . This contradiction completes the proof of (1.1.1).

Next, assume that X is Baire separable and metrizable, {Tt}tA is mixing and equation is a sequence of elements of A such that |tn| → ∞. From the definition of mixing it follows that for any non-empty open subsets U and V of X, equation for all sufficiently large equation Hence equation is dense in X × X. By Theorem U, equation is universal.

Theorem 2.13. Let X be a topological vector space, equation and A=(A1,..., Ak) ∈ ε (k, X) be a EBSk -tuple. Then the strongly continuous group equation is mixing. If additionally X is Baire separable and metrizable, equation is hereditarily hypercyclic.

Proof. Assume that equation is non-mixing. Then we can find nonempty open subsets U and V of X and a sequence equation in equation such that |zm| → ∞ as m → ∞ and equation for each equation Let Σ be the set of pairs (x, y) ∈ X × X for which there exists a sequence equation of elements of X such that xm → x and equation. According to Corollary 2.10, ker (A) × ker (A) ⊆ Σ. Since A=(A1,..., Ak) is a EBSk -tuple, ker (A) is dense in X and therefore Σ is dense in X × X. In particular, Σ meets U × V, which is not possible since equation for any equation This contradiction shows that equation is mixing. If X is Baire separable and metrizable, Proposition 1.1 implies that equation is hereditarily hypercyclic.

It is easy to see that if X is a Banach space and equation Moreover, each operator group of the shape equation is uniformly continuous. Hence, we get the following corollary of Theorem 2.13.

Corollary 2.14. Let X be a separable Banach space and (A1,..., Ak) ∈ L(X)k be a EBSk -tuple. Then equation is a hereditarily hypercyclic uniformly continuous group.

equation-sequences, equicontinuous sets and the class equation

Lemma 3.1. Let Y0 and Y1 be closed linear subspaces of a locally convex space Y such that Y0 ⊂ Y1 and the topology of Y1/Y0 is not weak. Then there is a sequence equation in Y′such that

(3.1.1) equation

(3.1.2) equation for each equation

(3.1.3) equation is uniformly equicontinuous.

Proof. Since the topology of Y1/Y0 is not weak, there exists a continuous seminorm equation on Y1/Y0 such that the closed linear space ker equation=equation-1 (0) has infinite codimension in Y1/Y0. Clearly the seminorm p on Y1 defined by the formula equation is also continuous and ker p has infinite codimension and contains Y0. In particular the space Yp=Y1/ker p endowed with the norm equation is an infinite dimensional normed space. Hence we can choose sequences equation in Y1 and equation in Y′p such that ||gn|| ≤ 1 for each equation and gn(yk + ker p)=δn,k for n, equation where δn,k is the Kronecker δ (every infinite dimensional normed space admits a biorthogonal sequence). Now let the functionals equation be defined by the formula hn(y)=gn(yk + ker p). The above properties of the functionals gn can be rewritten in terms of hn in the following way equation for any n, equation

Since any continuous seminorm on a subspace of a locally convex space extends to a continuous seminorm on the entire space [17,29], we can find a continuous seminorm q on Y such that q|Y1 = p .

Applying the Hahn–Banach theorem, we can find equation such that

equation

From the last two displays we have fn(yk)=δn,k, which implies (3.1.1) since yk ∈ Y1. From the inequality in the above display it follows that each |fn| is bounded by 1 on the unit ball W of the seminorm q. Since W is a neighborhood of zero in Y, condition (3.1.3) is satisfied. Since Y0 ⊆ ker p ⊆ ker q, from the inequality equation it follows that each fn vanishes on Y0. That is, (3.1.2) is satisfied.

Applying Lemma 3.1 with Y0=0 and Y1=Y, we obtain the following corollary.

Corollary 3.2. Let Y be a locally convex space, whose topology is not weak. Then there exists a linearly independent sequence equation in Y′ such that equation is uniformly equicontinuous and equation

Recall that a subset D of a locally convex space X is called a disk if D is bounded, convex and balanced (=is stable under multiplication by any equation The symbol XD stands for the space span (D) endowed with the norm being the Minkowskii functional of the set D. Boundedness of D implies that the topology of XD is stronger than the one inherited from X. A disk D in X is called a Banach disk if the normed space XD is complete. It is well-known that a sequentially complete disk is a Banach disk, see, for instance, [29]. In particular, a compact or a sequentially compact disk is a Banach disk. We say that D is a Banach s-disk in X if D is a Banach disk and the Banach space XD is separable.

Lemma 3.3. Let equation be an equation sequence in a locally convex space X. Then the set

equation

is a compact and metrizable disk. Moreover, K is a Banach s-disk and E=span equation is dense in the Banach space XK.

Proof. Let equation be endowed with the coordinatewise convergence topology. It is easy to see that Q is a metrizable compact topological space as a closed subspace of
equation Obviously, the map equation is onto. It is also easy to see that φ is continuous. Indeed, let p be a continuous seminorm on X, a ∈ Q and ε > 0. Since xn → 0, there is equation such that p(xn) ≤ ε for n > m. Let δ=ε(1 + p(x0) +... + p(xm))−1 and W={b ∈ Q : |aj − bj | < δ for 0 ≤ j ≤ m}. Then W is a neighborhood of a in Q and for each b ∈ W we have

equation

Taking into account that p(xn) < ε for n > m and |an − bn| < δ for n ≤ m, we obtain

equation

Using the definition of δ and inequalities ||a||1 ≤ 1, ||b||1 ≤ 1, we see that p(φ(b) − φ(a)) ≤ 3ε. Since a, p and ε are arbitrary, the map φ is continuous. Hence K is compact and metrizable as a continuous image of a compact metrizable space. Obviously K is convex and balanced. Hence K is a Banach disk (any compact disk is a Banach disk). Let us show that E is dense in XK. Take u ∈ XK. Then there is a ∈ equation such that

equation

Let || . || be the norm of the Banach space XK. Then for any equation
equation

Hence E is dense in XK and therefore XK is separable and K is a Banach s-disk.

Lemma 3.4. Let X be a separable metrizable topological vector space and equation be a linearly independent sequence in X. Then there exist sequences equation in X and equation in equation such that span equation is dense in X, gn(xk)=0 for n ≠ k and gn(xk) ≠ 0 for equation where equation

Proof. Let equation be a base of topology of X. First, we construct inductively sequences
equation in equation and equation in X such that for any equation

(b1) yk ∈ Uk;

(b2) gk (yk) ≠ 0, where equation

(b3) gk (yk)=0 if m < k.

Let g0=f0. Since f0 ≠ 0, there is y0 ∈ U0 such that f0(y0)=g0(y0) ≠ 0. This provides us with the base of induction. Assume now that n ∈ equation and yk, αk,j with j < k < n satisfying (b1– b3) are already constructed. According to (b2) and (b3), we can find αn,0,..., αn,n−1equation such that gn(ym)=0 for m < n, where equation Next since fj are linearly independent, gn ≠ 0 and therefore there is yn ∈ Un such that gn(yn) ≠ 0. This concludes the description of the inductive procedure of constructing sequences equation satisfying (b1–b3) for each equation

Using (b2) and (b3), one can easily demonstrate that there is a sequence equation in equation such that gn(xk)=0 for k ≠ m, equation . From (b2) and (b3) it also follows that gn(xn) 0 for each equation It remains to notice that according to (b1), equation is dense in X. Since equation ⊆ span equation =span equation we see that span equation is dense in X.

Lemma 3.5. Let equation Then there exist sequences equation and equation in X and X′ rspectively, such that

(3.5.1) equation is an equation-sequence in X;

3.5.2) the space E=span equation is dense in X;

(3.5.3) fk (xn)=0 if k ≠ n and fk (xk) ≠ 0 for each equation

(3.5.4) equation is uniformly equicontinuous.

Moreover, {fk} can be chosen from the linear span of any linearly independent uniformly equicontinuous sequence in X′.

Proof. According to Lemma 3.3, there exists a Banach s-disk K in X such that XK is dense in X. By Corollary 3.2, there is a linearly independent sequence equation in X′ such that equation is uniformly equicontinuous. Since XK is dense in X, the functionals equation on XK are linearly independent. Applying Lemma 3.4 to the sequence equation, we find that there exist sequences equation in XK and equation in equation such that E=span equation is dense in XK, hn(yk)=0 for n ≠k and hn(yn) ≠ 0 for equation where equation Let q be the norm of the Banach space XK. Consider fn=cnhn, where equation. Since equation is uniformly equicontinuous and equation we immediately see that equation is uniformly equicontinuous. Next, let equation Since xn converges to 0 in the Banach space equation -sequence in XK. Since the topology of XK is stronger than the one inherited from X, equation -sequence in X. Since span{ equation is dense in XK, it is also dense in XK in the topology inherited from X. Since XK is dense in X, span equation is dense in X. Finally since fn(xk)=cnbk hn(yk), we see that fn(xk)=0 if n ≠ k and fn(xn) ≠ 0 for any equation Thus conditions (3.5.1–3.5.4) are satisfied.

Lemma 3.6. Let X be a locally convex space, whose topology is not weak and Y be a locally convex space admitting an equation-sequence with dense span. Then there is T ∈ L(Y, X) such that T (X) is dense in Y.

Proof. Let equation be an equation -sequence in Y with dense span. By Corollary 3.2, there is a uniformly equicontinuous sequence equation in X′ such that equation ,. Consider the linear operator T : X → Y defined by the formula equation .Since {fn} is uniformly equicontinuous, there is a continuous seminorm p on X such that |fn(x)| ≤ p(x) for any x ∈ X and equation Since {yn} is an equation -sequence, Lemma 3.3 implies that the closed convex balanced hull Q of equation is compact and metrizable. Let q be the Minkowskii functional of Q (=the norm on YQ). It is easy to see that q(T x) ≤ p(x) for each x ∈ X. Hence T is continuous as an operator from X to the Banach space YQ. Since the latter carries the topology stronger than the one inherited from Y, T ∈ L(X, Y). Next, the inclusion equation , implies that T (X) contains the linear span of equation, which is dense in Y. Thus T has dense range.

Lemma 3.7. Let X be an infinite dimensional locally convex space and assume that there exist equation -sequences with dense span in both X and X′β. Then there exist sequences equation and equation in X and X′ respectively, such that.

(3.7.1) equation is an equation -sequence in X and equation is an equation -sequence in X′β;

(3.7.2) E=span equation is dense in X and F=span equation is dense in X′β;

(3.7.3) fk(xn)=0 if k ≠ n and fk (xk) ≠ 0 for each equation

If the original equation-sequence in X′β is uniformly equicontinuous, than we can also ensure that (3.7.4) equation is uniformly equicontinuous.

Proof. By Lemma 3.3, there exists a Banach s-disk K in X such that XK is dense in X. Fix an equation-sequence equation in X′β with dense span. Since any sequence of elements of a linear space with infinite dimensional span has a linearly independent subsequence with the same span, we, passing to a subsequence, if necessary, can assume that gn are linearly independent. Since XK is dense in X, the functionals equation on XK are linearly independent. Let D be the closed absolutely convex hull of the set equation in X′βBy Lemma 3.3, D is a Banach disk in X′β Applying Lemma 3.4 to the sequence equation , we find that there exist sequences equation in XK and equation in equation such that E=span equation is dense in XK, hn(yk)=0 for n ≠ k and hn(yn) ≠ 0 for equation where equation Let q be the norm of the Banach space XK and p be the norm of the Banach space X′D. Consider fn=cnhn, where equation. Since p(gn) ≤ 1 for each equation and equation , we immediately see that equation for any equation Hence equation is an equation -sequence in the Banach space X′ D. Since the topology of X′ D is stronger than the one inherited from X′β, equation is an equation -sequence in X′β. Similarly, equation is uniformly equicontinuous if equation is. Since the sequences equation have the same spans, the span of equation is dense in X′β. Next, let xn=bnyn, where bn=2−nq(xn)−1. Since xn converges to 0 in the Banach space XK, equation is an equation -sequence in XK. Since the topology of XK is stronger than the one inherited from X, equation is an equation - sequence in X. Since span equation =span equation is dense in XK, it is also dense in XK in the topology inherited from X. Since XK is dense in X, span equation is dense in X. Finally since fn(xk)=cnbkhn(yk), we see that fn(xk)=0 if n ≠ k and fn(xn) ≠ 0 for any equation Thus conditions (3.7.1–3.5.3) (and (3.5.4) if equation is uniformly equicontinuous) are satisfied.

Lemma 3.8. The class equation is stable under finite or countable products. Moreover, equation

Proof. It is clear that if the topology of one of the locally convex spaces Xj is not weak, then so is the topology of their product ΠX j . Thus the only thing we have to worry about is the existence of an equation-sequence with dense span. Let J be a finite or countable infinite set, equation for each j ∈ J and equation . Let for any j ∈ J, equation be an equation-sequence in Xj with dense span. For (j, n) ∈ equation let uj,n ∈ X be such that jth component of uj,n is xj,n, while all other components are 0. Clearly {uj,n : j ∈ J,equation} is a countable subset of X. It is easy to see that enumerating this subset by elements of equation, we get an equation-sequence in X with dense span. The proof of the second part of the lemma is even easier: one have just to add one vector (0, 1) to an equation-sequence with dense span in X=X × {0}, to obtain an equation-sequence with dense span in X × equation.

Proof of Theorem 1.3

Let equation By Lemma 3.5, there exist sequences equation and equation in X and X′ respectively satisfying (3.5.1–3.5.4). Uniform equicontinuity of {fn} is equivalent to the existence of a continuous seminorm p on X such that each |fn| is bounded by 1 on the unit ball {x ∈ X : p(x) ≤ 1} of p. Since {xn} is an equation-sequence in X, Lemma 3.3, absolutely convex closed hull K of {xn : equation} is compact and is a Banach disk in X. Let q be the norm of the Banach space XK. Then q(xn) ≤ 1 for each equation Let c=sup{p(x) : x ∈ K}. Compactness of K implies that c is finite. Clearly c > 0 and p(x) ≤ cq(x) for any x ∈ Y.

Lemma 4.1. Let α, β : equation be any maps and equation Then the formula

equation (4.1)

 

defines a linear operator on X. Moreover, the series equation converges in X for any x ∈ X anid equation and

equation for each x ∈ X and equation (4.2)

 

where ||a|| is the equation-norm of a.

Proof. Condition (3.5.4) implies that the sequence equation is bounded for any x ∈ X. Since {xn} is an equation-sequence and a ∈ equation, we see that the series in (4.1) converges for any x ∈ X and therefore defines a linear operator on X. Moreover, if p(x) ≤ 1, then |fk (x)| ≤ 1 for each equation and since q(xm) ≤ 1 for equation formula (4.1) implies that q(Tx) ≤ ||a||. Hence q(Tx)≤ ||a||p(x) for each x ∈ X. Then q(T2x) ≤ ||a|| p(Tx) ≤ c||a||q(Tx) ≤ c||a||2p(x). Iterating this argument, we see that

q(Tnx) ≤ cn−1||a||n p(x) for each equation

It follows that for any x ∈ X, the series equation is absolutely convergent in the Banach space XK (and therefore in X) for any equation. Naturally we denote its sum as ezT x. Moreover, using the above display, we obtain

equation

Thus (4.2) is satisfied.

Now fix a bijection equation By symbol ej we denote the element of equation defined by equation, where equation is the Kronecker delta. For equation , we also write equation. For each equationlet

equation

According to (3.5.1), (3.5.3) and (3.5.4), {εm} is a bounded sequence positive numbers. Next, we pick any sequenceequation of positive numbers such that

equation for any equation                (4.3)

 

and consider the operators equation defined by the formula

equation

From the estimates (4.3) it follows that the series defining Aj can be written as

equation with equation

Clearly Aj have shape (4.1) with equation.By Lemma 4.1, Aj are linear operators on X satisfying equation for any equation. Hence Aj are continuous as operators from X to the Banach space XK. Since XK carries a stronger topology than the one inherited from equation. Next, using the definition of Aj , it is easy to verify that equation for any equation and equation Indeed, for any equation, there is a unique equation such that equationIf either equation, from the definition of Aj and Ai it follows that equation. If equation from the same definition we obtain equation.From (3.5.2) it follows now that the operators equation are pairwise commuting. Next, for equation, the operator equation has shape (4.1) with equation, where equation is the equation-norm of z. By Lemma 4.1, the series equation converges pointwisely to a linear operator equation and

equationfor any equation             (4.4)

 

Exactly as for the operators Aj, we see that equation for any equation From (4.4), if equation uniformly on equationBy Proposition 2.8, equation is a uniformly continuous group and the map equation is holomorphic for any equation provided equation The proof will be complete if we show that the group equation is hereditarily hypercyclic. To this end consider the restrictions Bj of Aj to XK as bounded linear operators on the Banach space XK. Then Bj commute with each other as the restrictions of commuting operators. Let us show that the operator group equation on XK is hereditarily hypercyclic. By Corollary 2.14, it suffices to verify that equation is a EBSk -tuple. We already know that Bj are pairwise commuting. From the definition of Bj it is easy to see that ker equation contains equation .Using this fact it is easy to see that for any equation the set equationdefined by (2.2) contains

equation

Hence the space equation defined in (2.2) contains equation . By Lemma 3.3 equation is dense in XK. Hence B is an EBSK -tuple. Thus according to Corollary 2.14, equationis hereditarily hypercyclic. Since XK is dense in X and carries a topology stronger than the one inherited from X,equation is hereditarily hypercyclic. By Proposition 1.1, equation

Theorem 1.18: proof and applications

Proof of Theorem 1.18. Assume that T is non-mixing. Then we can choose non-empty open subsets U and V of X and a strictly increasing sequence equation of positive integers such that equation for any equation Let Σ be the set of equation for which there exists a sequence equation of elements of X such that equation According to Theorem 2.12, Σ contains equationSince Λ(T) is dense in X, we see that Σ is dense in X × X. Hence Σ intersects U × V, which is impossible since equation for any equation This contradiction shows that T is mixing. If X is Baire separable and metrizable, then by Proposition 1.1, T is hereditarily hypercyclic.

Extensions of the Salas theorem

It is easy to see that kerequation for any linear operator T. Thus Theorem 1.18 implies the following corollary.

Corollary 5.1. Let T be an extended backward shift on a topological vector space X. Then I + T is mixing. If additionally X is Baire, separable and metrizable, then I + T is hereditarily hypercyclic.

Recall that a backward weighted shift on equation is the operator T acting on the canonical basis equation as follows: equation where equation is a bounded sequence of non-zero numbers in K. Clearly any backward weighted shift is a generalized backward shift and therefore is an extended backward shift. Hence Corollary 5.1 contains the Salas theorem on hypercyclicity of the operators I + T with T being a backward weighted shift as a particular case. It is also easy to see that if X is a topological vector space and equationis surjective, then equation.Thus we obtain the following corollary.

Corollary 5.2. Let X be a topological vector space and equation be such that T (X)=X and ker* T is dense in X. Then I + T is mixing. If additionally X is Baire, separable and metrizable, then I + T is hereditarily hypercyclic.

We can further generalize Corollary 5.1 by means of the following observation.

Lemma 5.3. Let X be a topological vector space, k ∈ N, T ∈ L(X) and {aj}j∈N be a sequence in k such that the seriesequation of operators converges pointwise to a continuous linear operator S on X. Then equation

Proof. In order to prove the inclusion equation, it is enough to show that equationfor any n ∈ N Thus pick n ∈ N and let equation.Then equationand there is y ∈ X such that equation.For equationdenote equation.It is easy to see that equation, equationfor 2 ≤ j ≤ 2kn and equation. Consider the backward shift B on the space equation, which acts on the canonical basic vectors by the same rule: Be1=0 and equation for 2 ≤ j ≤ 2kn. Let Y =span{hj : 1 ≤ j ≤ 2kn} and consider the surjective linear operator equationdefined by equationforequation.It is easy to see that Y is T-invariant and equation,where equationthe restriction of T to Y. Since B2kn=0, we see that equationand therefore the restriction equationto Y is given by the formula equation, where P is the polynomial defined by equation.Let E =span{ej : 1 ≤ j ≤ kn}. Considering the matrix of the operator p(B), it is easy to see that equation.Using the intertwining relation equation, we see that equation.Since equation is the restriction to Y of the operator Tk +S, we see thatequation

The following result is an immediate consequence of Lemma 5.3 and Corollary 5.1.

Corollary 5.4 Let X be a topological vector space,equation and be a sequence in K such that the series equationof operators converges pointwise to a continuous linear operator S on X. Assume also that equation is an extended backward shift. Then the operator I+equation+S is mixing. If additionally X is Baire separable and metrizable, then I+equation+S is hereditarily hypercyclic.

It is straightforward to verify that any power of a generalized weighted shift is an extended backward shift. Thus the above corollary implies the next observation

Corollary 5.5 Let X be a topological vector space, T ∈ L(X) be a generalized backward shift and equation be a sequence in K such that equation, there is n ∈ N for which equation and the series equation converges pointwise to a continuous linear operator S on X. Then S is mixing. If additionally X is Baire separable and metrizable, then S is hereditarily hypercyclic.

An extension of the Hilden–Wallen theorem

Lemma 5.6 Let X be a Baire topological space, Y be a second countable topological space and equation be a sequence of continuous maps from X to Y. Let alsoΣ be the set of equationfor which there exists a sequence equation of elements of X such that equation and equation as equation. If Σ is dense in X×Y, then equationis hereditarily universal.

Proof. The density of Σ in X×Y implies that for any infinite set equation,condition (U3) from Theorem U is satisfied for the family equationand therefore this family is universal.

Hilden and Wallen [30] demonstrated that any backward weighted shift on equationis supercyclic. Many particular cases of the following proposition are known, see for instance [31]. We include it here in its full generality for the sake of completeness.

Proposition 5.7 Let X be a Baire separable metrizable topological vector space T ∈ L(X). Suppose also that T has dense range and dense generalized kernel. Then T is hereditarily supercyclic.

Proof. Since T has dense range, we have that Tk(X) is dense in X for each equation. Thus for any x ∈X there exists a sequence equationin X such that equationas equation.Fix a dense countable set equation.Since X is metrizable and B is countable, we can choose a sequence equation of positive numbers such that equationasequationfor eachequation.

Let now equationfor eachequationand Σ be the set of equationfor which there exists a sequence equation of elements of X such that equation and equation.For any equation,let equation.Then equation and equation for each equation. Hence equation. On the other hand, for any equation,we have equation for all sufficiently large k and therefore equation. Considering the constant sequence equation,we see that equation.

Finally, observe that Σ is a linear subspace of X×X. Hence equation.Since both B and ker *T are dense in X, we see that Σ is dense in X× X. By Lemma 5.6, for each infinite setequation.the family equationis universal. Hence T is hereditarily supercyclic.

Remarks on Theorem 1.18

Theorem 1.18 is reminiscent of the following criterion of hypercyclicity of Bayart and Grivaux [32] in terms of the unimodular point spectrum.

Theorem BG Let X be a complex separable infinite dimensional Banach space, T ∈ L(X) and assume that there exists a continuous Borel probability measure μ on the unit circle T such that for each Borel set equation with μ(A)=1, the space

equation

is dense in X. Then T is hypercyclic.

It is worth noting that in the case of operators on Banach spaces, neither Theorem 1.18 implies the result of Bayart and Grivaux, nor their result implies Theorem 1.18, see Examples 5.9, 5.10 and 5.12 below. Theorem 1.18 is also strictly stronger than Proposition 2.2 in the article [33] by Herrero and Wang, which is a key tool in the proof of the main result in reference [33] that any element of the operator norm closure of the set of hypercyclic operators on l2 is a compact perturbation of a hypercyclic operator. Namely, they assume that the span taken in (1) is dense taking only into account the z’s for which T -zI has closed range, and this allows them to use the Kitai criterion. We would like also to mention the following fact.

Proposition 5.8 Let X be a locally convex space and Y be a closed linear subspace of X such that X admits an l1-sequence with dense span and the topology of X/Y is not weak. Then there is a hereditarily hypercyclic operator T ∈ L(X) such that Ty=y for any y ∈ Y.

Proof. Since the topology of X/Y is not weak, Lemma 3.1 implies that there is a linearly independent uniformly equicontinuous sequence equation such that Y ⊆ ker gn for each n ∈ Z+ .By Lemma 3.5, we can find sequence equation and equation in X and X′ respectively, such that conditions (3.5.1–3.5.4) are satisfied and each fk belongs to span equation. Hence equation for any equation Uniform equicontinuity of { fk} and the fact that {xk} is an l1-sequence implies that the formula

equation

defines a continuous linear operator on X, which also acts continuously on the Banach space Xk, where K is the Banach disk being the closed convex balanced hull of equation. By Lemma 3.3, Xk is separable and equation is dense in Xk .It is also straightforward to verify that equation contains E, where equation is the restriction of T to Xk. Hence Tk is an extended backward shift on the separable Banach space Xk. By Corollary 5.1, I+Tk is hereditarily hypercyclic. Since Xk is dense in X and carries stronger topology,equationis hereditarily hypercyclic. Since equation for each equation, from the definition of T it follows that equation for each y ∈ Y. Thus I + T satisfies all required conditions.

Grivaux [34] proved that if Y is a closed linear subspace of a separable Banach space X such that X/Y is infinite dimensional, then there exists a hypercyclic T ∈ L(X) such that Ty=y for any y ∈ Y. This result is an immediate corollary of Proposition 5.8. If we apply Proposition 5.8 in the case when X is a separable Fréchet space, we obtain that whenever Y is a closed linear subspace of X such that X/Y is infinite dimensional and non-isomorphic to ω, there is a hereditarily hypercyclic T ∈ L(X) such that Ty=y for any y ∈ Y. It is also worth noting that any non-normable Fréchet space has a quotient isomorphic to ω, [29]. If X is a non-normable separable Fréchet space non-isomorphic to ω and Y is a closed linear subspace of X such that X/Y is isomorphic to ω, then there is no hypercyclic operator T ∈ L(X) such that Ty=y for each y ∈ Y. Indeed, assume that such a T does exist. Then equation , equationis a continuous linear operator with dense range from X/Y to X. Since X/Y is isomorphic to ω and ω carries the minimal locally convex topology, S is onto and therefore X is isomorphic to ω, which is a contradiction.

Example 5.9 Let equation be the complex Sobolev space equation which consists of the distributions f on T such that equation where the equation denotes the nth Fourier coefficient of f. Then the operator T acting on equation as equation satisfies the conditions of Theorem BG (and therefore is hypercyclic) and does not satisfy the conditions of Theorem 1.18.

Proof. For each equation is the onedimensional space spanned by the Dirac δ -function δz, which does not belong to the range of equation.Thus equationand T does not satisfy conditions of Theorem 1.18. On the other hand, equation is dense in equation for any set equationwhich is dense in T. Since any subset of T of full Lebesgue measure is dense, the conditions Theorem BG are satisfied with μ being the normalized Lebesgue measure.

Example 5.10 If equation is the complex Sobolev space equationwhich consists of the distributions f on T such that equation,then the operator equationacting on equation satisfies the conditions of both Theorem BG and Theorem 1.18.

Proof. For each equation and therefore the dense linear span of the set equation is contained in equation.Hence T satisfies the conditions of Theorem 1.18. As in the above example, equationis dense in equationfor any set equation which is dense in T and therefore T satisfies conditions of Theorem BG.

Example 5.11 If T is a quasinilpotent generalized backward shift on a separable complex Banach space, then I+T satisfies the conditions of Theorem 1.18 and does not satisfy the conditions of Theorem BG.

Proof. Since equationconditions of Theorem BG are not satisfied for the operator I+T. On the other hand, equationis dense and therefore I+T satisfies conditions of Theorem 1.18.

Examples of chaotic operators on a complex Hilbert space H which are not mixing are constructed in [35]. For such an operator, the linear span of the union of equation satisfying equationfor some equation is dense in H. This shows that the assumption of Theorem 1.18 cannot be relaxed into a weaker assumption like density of the linear span of the union of equation. Note also that the class of operators T for which equation is dense is closed under finite direct sums. Moreover, this class for operators acting on Banach spaces is closed under infinite c0-sums and lp-sums for equation . In particular, I+T is mixing, when T is a finite or countable c0-sum or lp- sum with equationof (possibly different) backward weighted shifts.

Now we describe another class of operators to which Theorem 1.18 applies. Let equation and equation be a Borel measurable map. Consider the integral operator equationdefined by

equation

It is straightforward to verify that equation is a compact linear operator on equation

Example 5.12 Assume thatΨ is continuous, strictly increasing equationand that equationalmost everywhere on [0.1]. Then equation acting on equation is hereditarily supercyclic and I+T is mixing.

Proof. Consider the sequence equationClearly equation is strictly decreasing and tends to zero as equation.One can easily verify that

equation

Using this equality it is straightforward to check that equation is dense in equation for each equation and that equation is dense in the entire space. Thus equation is dense in equation.That is, T is an extended backward shift. It remains to apply Proposition 5.7 and Corollary 5.1.

Universality of generic families

Theorem 6.1 Let X be a separable metrizable Baire topological space, Ω be a Baire topological space, A be a set, and for each equation let Ψa be a map from Ω into the set C(X, X) of continuous maps from X to X such that

equation(6.1)

 

equation (6.2)

 

Then

equation

Proof. Let equation be a sequence of non-empty open subsets of X, which form a basis of the topology of X. By Theorem U, equationis universal if and only if for each equationthere exists equation for which equation.Thus we have

equation

According to (18), the sets equationare open in Ω. Hence, the above display implies that U is a Gδ -subset of Ω. It remains to show that U is dense in Ω. Let

equation

Clearly U is the projection of U0 onto Ω. On the other hand, U0 is the set of universal elements of the family equation.Since the product of two Baire spaces, one of which is second countable, is Baire [36], Ω × X is Baire. Applying Theorem U, we see that U0 is dense in Ω × X. Since the projection onto Ω of a dense subset of Ω × X is dense, we get that U is dense in Ω.

We apply the above general result to two types of universality: hypercyclicity and supercyclicity.

Theorem 6.2 Let X be a separable metrizable Baire topological vector space, Ω be a Baire topological space and equationbe a map from Ω intoL(X)such that

equation(6.3)

 

Then the following conditions are equivalent:

equation(6.4)

 

equation(6.5)

 

Proof. Let equationandfequationforequationbe defined as equation.Applying Theorem 6.1, we see that (6.4) implies (6.5). Since the set of hypercyclic vectors of any hypercyclic operator is dense, (6.5) implies (6.4).

Theorem 6.3 Let X be a separable metrizable Baire topological vector space, Ω be a Baire topological space and equation be a map from Ω to L(X) such that (6.3) is satisfied. Then the following conditions are equivalent:

equation(6.6)

 

equation(6.7)

 

Proof. Let equation and equationfor equationbe defined as equation.Applying Theorem 6.1, we see that (6.6) implies (6.7). Since the set of supercyclic vectors of any supercyclic operator is dense, (6.7) implies (6.6).

Corollary 6.4 Let X be a separable Baire metrizable topological vector space, Ω be a Baire topological space and equation be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U, V of X, there exist equation, equation and equation such that equation,where equation is defined in (1.1). Then the set of α ∈ Ω for which Tα is hypercyclic is a dense G -subset of Ω.

Proof.Let equationbe such that equation.By Theorem 2.12 (x, y) is in the closure of the set equation.Hence equationis in the closure of the set equation By the assumptions of the corollary, the last set is dense in Ω × X × X. It remains to apply Theorem 6.2.

The following supercyclicity analog of the above corollary turns out to be much easier.

Proposition 6.5 Let X be a Baire separable metrizable topological vector space, Ω be a Baire topological space and equation be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U and V of X, there exist equation, equation and n ∈ N such thatequation and equation. Then the set of equation for which Tα is supercyclic is a dense Gδ -subset of Ω.

Proof. Let equationbe such that there exists n ∈ N for which equationand equation.Let equation be such that equationFor each equationconsider equation.Then equation as equation.Moreover, equation for any equation.Thus for all sufficiently large m, equation.Since W, U and V were arbitrary, the set equation is dense in Ω × X × X. It remains to apply Theorem 6.3.

The obvious inclusion equationand the fact that a map equationsatisfies (6.3) if and only if the map equationsatisfies (6.3) imply the following corollary of Corollary 6.4 and Proposition 6.5.

Corollary 6.6 Let X be a Baire separable metrizable topological vector space, Ω be a Baire topological space and equation be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U and V of X, there exist equation, x ∈ U and y ∈ V such that equation.Then the set of equation for which equation is supercyclic and equationis hypercyclic is a dense Gδ-subset of Ω.

Now we apply the above general results. In particular, we shall prove Theorems 1.19 and 1.21. For a while we shall assume that X and Y are two infinite dimensional Banach spaces, equation is a continuous bilinear form separating points of X and of Y. That is, for each non-zero x ∈ X, there is y ∈ Y satisfying equation and for each non-zero y ∈ Y, there is x ∈ X such that equation. In particular,b is a dual pairing between X and Y. Recall that the injective norm on the tensor product equationis defined by the formula

equation

The completion of equation with respect to this norm is called the injective tensor product of X and is Y and denoted equation.It is again a Banach space. For each equation we consider the linear operators equationand equationdefined by the formulae

equation (6.8)

 

Clearly equation and equation are bounded linear operators from equation endowed with the injective norm into the Banach spaces L(X) and L(Y) respectively. Hence they admit unique continuous extensions to equation , which we again denote by equation and. Note that equation is the dual operator of equation with respect to the dual pairing b:

equation(6.9)

 

Let now M be a linear subspace ofequationwhich we suppose to be endowed with its own F-space topology stronger than the one inherited fromequation.Suppose also that X ⊗Y is a dense linear subspace of M and let equation be the closure in M of the set equation.Remark that (6.9) implies that equation if and only if equation. Thus nilpotency of equation is equivalent to nilpotency of equation and this implies that equation coincides with the closure in M of the set equation
Proposition 6.7 If the Banach space X is separable, then the set

equation

is a dense Gδ subset of equation.If X and Y are both separable, then the set

equation

is a dense Gδ subset of equation.

Proof. Let W be a non-empty open subset of equation and U1, U2 be nonempty open subsets of X. Pick equation. By the definition of equation,there exists equation such that equation and equationis nilpotent. Let equation be such that equation.Clearly the space

equation

has finite codimension in Y, while equation has finite codimension in X. Since b is a dual pairing of X and Y, we can find equation and equationsuch that equation.Consider the element

equation

and let equation for equation.From the above properties of fj and uj it immediately follows that the range of equation is contained in the range Q of equationand that the restrictions of equation to Q coincide. Hence equation for any equationThus all T equation are nilpotent and therefore all equation belong to equation.Since W is open in equationwe can pick equationclose enough to zero to ensure that equation. Now from the definition of equation .it immediately follows that equation, equation.Thus

equation

Summarizing the above, we have found equation such that equation.Using the fact that equationcarries a topology stronger than the one defined by the injective norm, we see that the map equation from equation to L(X) satisfies (6.3). By Corollary 6.6, the set of equation for which equation is supercyclic and equation is hypercyclic is a dense Gδ subset of equation. The proof of the first part of the proposition is complete.

Applying the first part of the proposition to (Y, X) instead of (X, Y) with b(x,y) replaced by b(y,x), we obtain that if Y is separable, then the set of equation for which equation is supercyclic and equation is hypercyclic is a dense Gδsubset of equation. The second part now follows from the first part and the fact that the intersection of two dense Gδ subsets of a complete metric space is again a dense Gδ set.

Proof of Theorems 1.19 and 1.21

Let us consider the case where Y=X′, b(x,y)=y(x) and equation.In this case the map equation is an isometry and equation is exactly the operator norm closure of the set of finite rank nilpotent operators. Moreover, taking into account that equation for any equation, we see that Theorem 1.19 is an immediate corollary of Proposition 6.7 for this specific choice of M and b.

Recall that the projective norm on the tensor product X ⊗ Y of the Banach spaces X and Y is defined by the formula

equation

where the infimum is taken over all possible representations of equation as a finite sum equation.If we consider the case equation and equationthe completion of equation with respect to the projective norm, then equation is exactly the set of nuclear quasinilpotent operators, and we see that Theorem 1.21 is an immediate corollary of Proposition 6.7 for this choice of M and b.

Proof of Theorem 1.6

Let X ∈ M By Lemma 3.5, there exist sequences equation and equation in X and X′ respectively such that conditions (3.5.1–3.5.4) are satisfied. Since{fn} is uniformly equicontinuous, we can pick a nonzero continuous seminorm p on X such that equation for any equation whenever equation. By Lemma 3.3, the closed balanced convex hull K of equation a Banach s-disk. That is, the Banach space Xk is separable. It is also clear from Lemma 3.3 that any equationhas shape equation for some equation and equation,where ||x|| is the norm of X in the Banach space Xk. Consider the bilinear form on X × l1 defined by the formula

equation

It is easy to see that b is well-defined, continuous and equation for any x ∈ X and a∈ l1. Moreover, b separates points of Xk.. and l1. Indeed, let x ∈ Xk., x ≠ 0. Pick α ∈ l1 such that equation.Since x ≠ 0, there is equationsuch that equation Using (3.5.3), we see that equation, where equation is the standard basis in l1. Similarly, let equation Then there is equationfor which am ≠ 0 and therefore, according to (3.5.3), equation Thus b separates points of Xkand l1. Let equation be the projective tensor product of the Banach spaces Xk and l1 and equation be the closure in M of the set of equation, for which the operator equation defined in (6.8) is nilpotent. By Proposition 6.7, the set equation is a dense Gδ subset of equation. In particular, we can pick equation such that equation and equation are hypercyclic. Using the theorem characterizing the shape of elements of the projective tensor product [17], we see that there exist bounded sequences equation and equation in Xk and l1 respectively and equationsuch that equation . Then the operators equation and equation act according to the following formulae on Xk and l1 respectively:

equation

Using boundedness of {yn} in Xk and {wn} in l1, summability of equation and the definition of b, we see that the right-hand side of the first equality in the above display defines a continuous linear operators T : X → X, taking values in Xk. Since the restriction equation of I+T to Xk is hypercyclic on Xk, Xk is dense in X and Xk carries the topology stronger than the one inherited from X, we see that I+T is hypercyclic (any hypercyclic vector for equation is also hypercyclic for I+T). Pick a hypercyclic vector a for equation. Since a ≠ 0 and b separates points of l1 , the functional b(⋅, a) is non-zero. Since a is hypercyclic for equation , we can pick a strictly increasing sequence equation of positive integers such that equation for any equation. Then

equation

Let equation and equation . Clearly U and V are non-empty open subsets of equation since the functional b(⋅,a) is non-zero). Moreover, from the above display it follows that equation for each equation. Hence I+T is non-mixing. Since I+T is hypercyclic, the proof of Theorem 1.6 is complete.

Proof of theorem 1.15

Let X be an infinite dimensional locally convex space, such that both X and equation admit l1-sequences with dense span. By Lemma 3.7, there exist sequences equation and equation in X and X' respectively, satisfying (3.7.1–3.7.3). By Lemma 3.3, the closed balanced convex hulls K and D of {xn: n ∈ equation} and {fn : n ∈equation} are Banach s-disks in X and equation respectively. Moreover, XK is dense in X and XD′ is dense in equation . Since D is β-compact, it is also σ (X′,X) -compact. Hence the seminorm equation on X is continuous with respect to the Mackey topology equation. Clearly each equation is bounded by 1 on equation and therefore {fn : n ∈equation} is uniformly equicontinuous. By Lemma 3.7, we can assume that the same holds true if the original l1-sequence in equation is uniformly equicontinuous. Assume for time being that either X carries the Mackey topology τ or the original l1-sequence in equation is uniformly equicontinuous. Then {fn : n ∈equation} is uniformly equicontinuous.

Consider the bilinear form on X x X′ defined by the formula β (x, f )= f (x) Clearly β separates points of X and X′ and β is separately continuous on equation. Since XK is dense in X and equation is dense in equation , the bilinear form equation , being the restriction of β to equation , separates points of XK and equation . Moreover separate continuity of β implies separate continuity of b and therefore continuity of b on equation by means of the uniform boundedness principle (every separately continuous bilinear form on product of Banach spaces is continuous). Let equation , for which the operator Tξ defined in (6.8) is nilpotent. By Proposition 6.7, the set equation is a dense Gδ subset of equation. In particular, we can pick equation such that equation and equation are hypercyclic. Using once again the theorem characterizing the shape of elements of the projective tensor product, we see that there exist bounded sequences equation and equation in XK and equation respectively and λ ∈ l1 such that equation. Then the operators Tξand Sξ act according to the following formulae on XK and equation respectively:

equation

where we used the specific shape of our bilinear form. Using boundedness of {yn} in XK and {gn} in equation and summability of equation , we see that the right-hand sides of the equalities in the above display define linear operators T and S on X and X′, taking values in XK and equation respectively. Since equation is uniformly equicontinuous and {gn} is bounded in equation, equation is also uniformly equicontinuous. It follows that T is continuous as an operator from X to XK and therefore T ∈ L(X). It is also easy to verify that S = T′ and therefore equation Since the restriction I +Tξ of I+T to XK is hypercyclic on XK, XK is dense in X and XK carries the topology stronger than the one inherited from X, we see that I+T is hypercyclic (any hypercyclic vector for I +Tξ is also hypercyclic for I+T). Similarly = equation is hypercyclic on equation (any hypercyclic vector for I + Sξ is also hypercyclic for I+S). Hence I+T is a dual hypercyclic operator. In order to complete the proof of Theorem 1.15, it remains to consider the case when X carries the weak topology σ =σ (X,X′) . By the already proven part of Theorem 1.15, there is a dual hypercyclic operator R on Xτ. Since equation , R ∈ L(X). Since τ is stronger than σ, R is hypercyclic on X= Xσ . Hence R is dual hypercyclic on X. The proof of Theorem 1.15 is complete.

Generic bilateral weighted shifts

For each equation , Tw stands for the bounded linear operator acting on equation , 1≤ p < ∞ or equation , defined on the canonical basis equation by

equation

If additionally wn≠ 0 for each equation, the operator Tw is called the bilateral weighted shift with weight sequence w. Cyclic properties of bilateral weighted shifts have been intensely studied. Hypercyclicity and supercyclicity of bilateral weighted shifts were characterized by Salas [12,31] in terms of the weight sequences. It was observed in [37], Proposition 5.1 that the Salas conditions admit a simpler equivalent form:

Theorem B Let T be a bilateral weighted shift with weight sequence w acting on equation with 1≤ p < ∞ or equation . Then T is hypercyclic if and only if for any equation
equation

and T is supercyclic if and only if for any equation (6.10)

equation (6.11)

 

where equation for equation with a ≤ b .

We address the issue of hypercyclicity and supercyclicity of generic bilateral weighted shifts in the Baire category sense. For each c > 0 let

equation

Clearly Bc endowed with the coordinatewise convergence topology is a compact metrizable topological space.

Theorem 6.8 Let 1≤ p < ∞ . For each c > 1 the set of w∈ Bc for which Tw acting on equation is hypercyclic is a dense Gδ- subset of Bc. For each c > 0 the set of w∈ Bc for which Tw is supercyclic and I+Tw is hypercyclic is a dense Gδ-subset of Bc.

It is worth noting that if wn=0 for some n ∈ equation then the range of the operator Tw is not dense and therefore Tw can not be supercyclic. Thus any Tw for w from the dense Gδ -sets in the above theorem are indeed bilateral weighted shifts. Recall that Tw is compact if and only if w∈ equation. For compact bilateral weighted shifts we can replace the coordinatewise convergence topology on the space of weights by stronger topologies.

Theorem 6.9 Let 1≤ p < ∞ . Let E be a linear subspace of equation carrying its own F -space topology stronger than the one inherited from equation and such that the space ψ(equation) of sequences with finite support is densely contained in E. Then the set of w∈E for which Tw acting on equation is supercyclic and I+Tw is hypercyclic is a dense Gδ-subset of E.

 

Remark 6.10 As a corollary of the above theorem we obtain that the set of weights w∈ equation for which Tw acting on equation is supercyclic and I+Tw is hypercyclic is a dense Gδ -subset of the Banach space equation.

It is easy to see that the dual of Tw acting on equation for 1< p <∞ acts on equation defined by equation according to the formula

equation

Considering the isometry U on equation defined by equation for each n ∈equation, we see that equation where equation for any n ∈equation. Thus equation is hypercyclic or supercyclic if and only if ′ (acting on equation is. In the case p =2, the Hilbert space adjoint Tw* acts on equation in a similar way equationand is unitarily similar to equation with a diagonal unitary operator providing the similarity. Thus the cyclicity properties of equation′ and Tw* are the same.

Taking into account the fact that the map equation is a homeomorphism of Bc onto itself for each c > 0, we immediately obtain the following corollary of Theorem 6.8.

Corollary 6.11 For each c>1 the set of w∈Bc for which both Tw and equation acting on equation are hypercyclic is a dense Gδ-subset of Bc. For each c > 0 the set of w∈ Bc for which both Tw and equation are supercyclic and both I + Tw and I +equation′ are hypercyclic is a dense Gδ-subset of Bc.

Similarly the next corollary follows from Theorem 6.9.

Corollary 6.12 Suppose that the space E from Theorem 6.9 satisfies the additional symmetry condition that equation is an invertible continuous linear operator on E for some k∈ equation. Then set of w∈ E for which both Tw and equation acting on equation are supercyclic and both I+Tw and I +equation are hypercyclic is a dense Gδ -subset of E.

Applying the above corollary to weighted c0-spaces with symmetric weight sequence yields the following result.

Corollary 6.13 Let equation be any sequence of positive numbers. Then there exists w∈ equation such that |wn|≤a|n| for each n ∈ equation, Tw and equation acting on equation are supercyclic and I+Tw and I +equation are hypercyclic

We conclude this section by proving Theorems 6.8 and 6.9.

Proof of Theorem 6.8. It is straightforward to verify that the maps equation and equation from Ω= Bc into equation satisfy (6.3). Pick a non-empty open subset U of Bc and non-empty open subsets V and W of equation.

Case c > 1: By definition of the topology of Bc, there exist w∈U and a positive integer m such that wk=c for k > m, wm=c-1 for k <-m and wk ≠ 0 for −m ≤ k ≤ m . According to Theorem B, the bilateral weighted shift Tw is hypercyclic. Hence we can choose x∈V and equation such that equation. Thus equation and therefore (6.4) is satisfied. By Theorem 6.2, the set of w∈ Bc for which Tw is hypercyclic is a dense Gδ -subset of Bc.

Case c > 0: As above, there exist w∈ U and a positive integer m such that wk=c for k > m, wk=c/2 for k <-m and wk ≠ 0 for −m ≤ k ≤ m. According to Theorem B, the bilateral weighted shift Tw is supercyclic Hence we can choose x∈ V, equation and equation such that equation . Thus equation and therefore (6.6) is satisfied. By Theorem 6.3, the set of w∈ Bc for which Tw is supercyclic is a dense Gδ-subset of Bc.

Finally, we can pick m ∈ equation, w∈ U, x∈ V, and y∈ W, such that wm= 0, wm ≠ 0 for k > m, x and y have finite supports and xk = yk = 0 for k > m. It is straight forward to check that

equation

By Corollary 6.6, the set of w∈ Bc for which I+Tw is hypercyclic is a dense Gδ-subset of Bc.

Proof of Theorem 6.9. We use the same notation as in the proof of Theorem 6.8. Since equation is dense in equation in E, we can choose w′∈U , x′∈V , y′∈W and a positive integer m such that x′k = yk′ =wk′ = 0 if |k |>m . Choosing positive numbers εn for n >-m small enough, we can ensure that the series equation converges in E to w′′∈E , with w = w′ + w′′∈U and equation for k>-m. Clearly equation is a closed invariant subspace of Tw and x,y ∈F. Moreover, the restriction R of Tw to F is isometrically similar to the backward weighted shift on lp with weight sequence equation By the Hilden and Wallen theorem [30] and the Salas theorem [12], R is supercyclic and I + R is hypercyclic. Since equation and equation we see that the open subsets V ∩F and W∩F of F are non-empty. Since R is supercyclic, we can choose x∈V ∩F , equation and equation such that equation . Thus equation and therefore (6.6) is satisfied. By Theorem 6.3, the set of w∈E for which Tw is supercyclic is a dense Gδ-subset of E. Finally, since I + R is hypercyclic, we can choose x∈V ∩F and equation such that equation . Thus equation and therefore (6.4) is satisfied for the map equation . By Theorem 6.2, the set of w∈E for which I+Tw is hypercyclic is a dense Gδ-subset of E.

Mixing Operators on Spaces with Weak Topology

In this section we shall prove Theorem 1.8. In order to do so we need a characterization of linear maps with no non-trivial finite dimensional invariant subspaces. The underlying field plays no role in this linear algebraic statement, so, for sake of generality we formulate and prove it for linear maps on linear spaces over an arbitrary field.

Linear maps without finite dimensional invariant subspaces

Throughout this subsection k is a field, X is a linear space over k, T: X→X is a k -linear map, P=k[z] is the space of polynomials on one variable, P=k(z) is the space of rational functions and M is the operator on P of multiplication by the argument:

equation

We denote equation and consider the degree function equation extending the conventional degree of a polynomial. We set deg (0)= −∞ and let deg (p / q)= deg p − deg q , where P and q are non-zero polynomials and the degrees in the right hand side are the conventional degrees of polynomials. Clearly this function is well-defined and is a grading on P, that is, it satisfies the properties:

equation for any equation ;

 

equation for any equation ;

 

(d#) if equation and equation , then equation.

Note that if p is a non-zero polynomial, then deg P is the usual degree of P.

As usual, a linear subspace E of X is called T - invariant if equation and it is called T- biinvariant if equation and equation . The following lemma is a key ingredient in the proof of Theorem 1.8.

Lemma 7.1 Let T be a linear operator on a linear space X with no non-trivial finite dimensional invariant subspaces and let L be a finite dimensional subspace of X. Then there exists equation such that p(T)(L)∩L = {0} for any equation with deg p ≥ n0 .

In order to prove the above lemma, we need some preparation.

Lemma 7.2 Let T be a linear operator on a linear space X. Then T has no non-trivial finite dimensional invariant subspaces if and only if p(T) is injective for any non-zero polynomial P.

Proof. If p is a non-zero polynomial and p(T) is non-injective, then there is non-zero x ∈ X such that p(T)x=0. Let k=deg p. It is straightforward to verify that equation is a nontrivial finite dimensional invariant subspace for T. Assume now that T has a non-trivial finite dimensional invariant subspac L. Let p be the characteristic polynomial of the restriction of T to L. By the Hamilton– Cayley theorem p(T) vanishes on L. Hence p(T) is non-injective.

Definition 7.3 For a linear operator T on a linear space X we say that vectors x1,…, xn in X are T - independent if for any polynomials p1,…, pn, the equality equation implies pj= 0 for 1≤ j ≤ n . Otherwise, we say that x1,…,xn are T - dependent. A set A ⊂ X is called T - independent if any pairwise different vectors equation are T - independent.

For any non-zero x ∈ X, we define

equation

Lemma 7.4 Let T be a linear operator on a linear space X, x ∈ X \{0} and F(T, x)be the space defined in (29). Then F(T, x) is a T -biinvariant linear subspace of X.

Proof. Let y, u∈ F(T, x) and t,s∈k . Then we can pick equation and equation such that equation and equation . Hence equation. Since p1 and p2 are non-zero, the polynomial p1 p2 is also non-zero and we have equation and therefore F(T, x) is a linear subspace of X. Clearly equation , where equation. Hence Ty∈ F(T, x), which proves the T-invariance of F(T, x). Assume now that w∈X and Tw∈ F(T, x). Thus we can pick equation and equation such that equation . Hence equation, where equation , and therefore w∈ F(T, x). That is, F(T, x) is T-biinvariant.

By the above lemma, we can consider linear operators, being restrictions of T to the invariant subspaces F(T, x). The following lemma describes these restrictions in the case when T has no nontrivial finite dimensional invariant subspaces.

Lemma 7.5 Let T be a linear operator on a linear space X with no non-trivial finite dimensional invariant subspaces. Let also x ∈ X \{0} and F(T, x) be defined in (7.1). For each y∈ F(T, x) and any equation, equation satisfying equation , we write rx,y= q/p. Then the rational function rx,y does not depend on the choice of equation and equation satisfying equation, the map equation from F(T, x) to P is linear and equation for any y∈ F(T, x).

Proof. Let y∈ F(T, x) and equation, equation be such that equation and equation. Hence equation and therefore equation. Since x ≠ 0, Lemma 7.2 implies that equation, or equivalently, q/p=q1/p1. Thus q/p does not depend on the choice of equation and equation satisfying equation . In particular, the map equation from F(T, x) to P is well defined. Next, let y, u∈ F(T, x) and t,s∈k . Then we can pick equation and equationsuch that equation and equation. Hence equation. It follows that

equation

and the linearity of the map equation is also verified. It remains to show thatequation . Clearly equation. Since equation, we have equation. Hence equation.

Lemma 7.6 Let T be a linear operator with no non-trivial finite dimensional invariant subspaces acting on a linear space X and A be a T -independent subset of X. Then for each x∈A ,

equation

where the spaces F(T, u) are defined in (7.1).

Proof. Assume that the intersection in the above display contains a non-zero vector u. Since u∈F(T, x) , there exist equation and equation such that equation . Since u ≠ 0 and p ≠ 0 , according to Lemma 7.2, p(T)u ≠ 0 . It follows that q ≠ 0 . On the other hand, since u is a non-zero element of the span of the union of F(T, Y) for y∈A\{x} , there exist pairwise different equationand equation such that equation . Pick equation and equation for which equation for 1≤ j ≤ n . Since uj ≠ 0 and pj ≠ 0 , from Lemma 7.2 it follows that equation and therefore equation . Consider the polynomials equation for 1≤ j ≤ n . Then

equation

Taking into account that equation , we obtain

equation

Since the polynomials equation and equation are non-zero, the last display contradicts the T -independence of A, since equation are pairwise different elements of A.

Proof of Lemma 7.1. Clearly there exists in L a maximal T -independent subset equation (since T -independence implies linear independence and L is finite dimensional, A is a finite set). It follows from the maximality of A that L is contained in the sum of F(T, x j ) for 1≤ j ≤ k . The last sum is direct according to Lemma 7.6:

equation

Thus any x ∈ N can be uniquely presented as a sum equation, where equation . Using Lemmas 7.5 and 7.4, we can consider the linear operator

equation

According to Lemmas 7.4 and 7.5 we also have that N is a T-biinvariant subspace of X and equation for 1≤ j ≤ k . For each x∈N , let

equation

Clearly δ (0)= −∞ and δ (x) ∈ equation for each x∈N \{0} . Let also

equation

Using the fact that L is finite dimensional, we will show that Δ+ and Δ- are finite. Indeed, assume that either Δ+ = +∞ or Δ- = −∞ . Then there exists a sequence equation of non-zero elements of L such that the sequence equation is strictly monotonic. For each l we can pick equation such that equation . Then there is equation such that the set equation is infinite. It follows that the degrees equation for equation are pairwise different. Property (d3) of the degree function implies that the rational functions equation for equation are linearly independent. Hence the infinite set equation is linearly independent in X, which is impossible since all u1 belong to the finite dimensional space L. Thus Δ+ and Δ are finite.

Now let equation and m= deg p . From (d1) and the equality equation, we immediately get that equation for each x∈N . Therefore, equation. In particular, if m> Δ+ − Δ , then

equation

Thus δ (u) >δ (v) for any non-zero u∈P(T)(L) and v∈L , which implies that equation whenever equation. Thus the number equation satisfies the desired condition.

Proof of Theorem 1.8

The implications (1.8.4)⇒(1.8.3)⇒(1.8.2) are trivial. Assume that T is transitive and T′ has a non-trivial finite dimensional invariant subspace. Then T has a non-trivial closed invariant subspace of finite codimension. Passing to the quotient by this subspace, we obtain a transitive operator on a finite dimensional topological vector space. Since there is only one Hausdorff linear topology on a finite dimensional space, we arrive to a transitive operator on a finite dimensional Banach space. Since transitivity and hypercyclicity for operators on separable Banach spaces are equivalent [3], we obatin a hypercyclic operator on a finite dimensional Banach space. On the other hand, it is well known that such operators do not exist, see, for instance, [7]. This proves the implication (1.8.2)⇒(1.8.1) . It remains to show that (1.8.1) implies (1.8.4).

Assume that (1.8.1) is satisfied and (1.8.4) fails. Then there exist non-empty open subsets U and V of X and a sequence equation of polynomials such that equation and equation for each equation . Since X carries weak topology, there exist two finite linearly independent sets equation and equation in X′ and two vectors equation and equation such that equation and equation , where

equation

Let equation . Since T′ has non non-trivial finite dimensional invariant subspaces, by Lemma 7.1, for any sufficiently large l , equation. For such an l, the equality equation together with the injectivity of equation , provided by Lemma 7.2, and the definition of L imply that the vectors equation are linearly independent. Hence there exists a vector u ∈ X such that

equation

Since equation, the last display implies that equation and equation. Hence equation contains equation and therefore is non-empty. This contradiction completes the proof.

Spaces without Supercyclic Semigroups equation

We shall prove Theorem 1.7 and show that on certain topological vector spaces there are no strongly continuous supercyclic semigroups equation. In this section by the dimension dimX of a vector space X we mean its algebraic dimension (=the cardinality of the Hamel basis). Symbol c stands for the cardinality of continuum: equation. The next theorem is the main result of this section.

Theorem 8.1 Let X be an infinite dimensional locally convex space such that either in X or in equation there are no compact metrizable subsets whose linear span has dimension c. Then there are no strongly continuous supercyclic semigroups equation on X.

The above theorem immediately implies the following corollary.

Corollary 8.2 Let X be an infinite dimensional locally convex space such that dim X < c or equation . Then there are no strongly continuous supercyclic semigroups equation on X.

We prove Theorem 8.1 at the end of this section. First, we shall prove Theorem 1.7 by means of application of Theorems 1.8 and 8.1.

Proof of Theorem1.7. Let T ∈L(ω) be such that T′ has no nontrivial finite dimensional invariant subspaces and equation be a sequence of polynomials such that equation as equation. Since the topology of ω is weak, Theorem 1.8 implies that for each non-empty open subsets U and V of ω, equation for all sufficiently large l. Hence equation is dense in ω×ω. By Theorem equation is universal. It remains to show that there are no strongly continuous supercyclic semigroups equation on X. Recall that equation and equation . Thus Corollary 8.2 implies that there are no supercyclic strongly continuous operator semigroups equation on ω.

The rest of the section is devoted to the proof of Theorem 8.1. We need some preparation.

Lemma 8.3 Let X be a finite dimensional topological vector space of the real dimension >2. Then there is no supercyclic strongly continuous operator semigroup equation on X.

Proof. As well-known, any strongly continuous operator semigroup equation on equation has shape equation, where equation such that the semigroup equation is supercyclic and n ≥ 3 if equation, n ≥ 2 if equation. Since the operators equation are invertible and commute with each other, Proposition G implies that the set W of universal elements for the family equation is dense in equation. On the other hand, for each c > 0 and any x ∈ equation, from the restrictions on n it follows that the closed set equation is nowhere dense in equation (one can use smoothness of the map equation to see that the topological dimension of equation is less than that of equation . Hence, each x∈W is universal for the family equation for any c > 0 . Now if (a,b) is a finite subinterval of (0,∞) , it is easy to see that the family equation contains equation for a sufficiently large c > 0. It follows that for each x ∈ W the set equation is dense in equation . Taking into account that (a,b) is arbitrary and W is dense in equation, we see thatequation is dense in equation . Applying Theorem 6.3, we see that for a generic t ∈equation in the Baire category sense the operator etA is supercyclic. This contradicts the well-known fact that there are no supercyclic operators on finite dimensional spaces of real dimension >2.

The following lemma appears as Lemma 2 in reference [38]. It is worth noting that under the Continuum Hypothesis its statement becomes trivial.

Lemma 8.4 Let (M, d) be a separable complete metric space and X be a topological vector space. Then for any continuous map f :M →X , the algebraic dimension of spanf(M) is either finite or countable or continuum.

We use the above lemma to prove the following dichotomy.

Lemma 8.5 Let equation be a strongly continuous operator semigroup on a topological vector space X and x ∈ X. Then the space equation is either finite dimensional or has dimension c.

Proof. From Lemma 8.4 it follows that dimC(x) is either finite or equation or c. Thus it suffices to rule out the case equation

Assume that equation . Restricting the operators Tt to the invariant subspace C(x) , we can without loss of generality assume that C(x) = X . Thus equation and therefore X is the union of an increasing sequence equation of finite dimensional subspaces. For each ε > 0 let equation . First, we shall show that each Xε is finite dimensional. Let ε > 0, 0 <α <ε and equation for equation. Clearly An are closed subsets of the interval [α ,ε ] and equation since X is the union of En. By the Baire category theorem equation there is such that An has non-empty interior in [α ,ε ] . Hence we can pick a,b ∈ equation such that α ≤ a < b ≤ε and Ttx∈En for any t∈[a,b] . We shall show that equation . Assume, it is not the case. Then the number equation belongs to [b,∞) . Since the set equation is closed, equation. Since [a,b] is uncountable and the span of equation is finite dimensional, we can pick equation and equation such that

equation

Since equation by definition of c, we can pick t equation such that equation. Since t > c ≥ tn , formula (8.1) implies that

equation

because equation for equation . This contradiction proves that equation. . Thus Xε is finite dimensional for each ε > 0. Since equation, it follows that Tt has finite rank for any t > 0.

Now assume that t > 0. Since Tt has finite rank, equation is a closed subspace of X of finite codimension. It is also clear that Ft is Ts-invariant for each s ∈equation Passing to quotient operators, Ss ∈L(X /Ft ), equation, we arrive to a strongly continuous semigroup equation on the finite dimensional space X / Ft . Hence there is equationsuch that equation for any s ∈ equation. Thus each Ss is invertible. Since each Ss is a quotient of Ts, we see that

equation

It follows that Tt for t > 0 have the same rank k ∈ equation. Passing to the limit as t → 0, we see that the identity operator I=T0 is the strong operator topology limit of a sequence of rank k operators. Hence rk I ≤ k . That is, X is finite dimensional. This contradiction completes the proof.

Lemma 8.6 :

Let X be a topological vector space in which the linear span of each metrizable compact subset has dimension <c. Then for any strongly continuous operator semigroup equation on X and any x ∈ X, the space equation is finite dimensional.

 

Proof: Let equation be a strongly continuous operator semigroup on X and x ∈ X. Strong continuity of equation implies that for any equation, the set equation is compact and metrizable. Hence equationfor any equation, where equation . Since the sum of countably many cardinals strictly less than c is strictly less than c, we see that

equation

By Lemma 8.5, equation is finite dimensional.

Proposition 8.7: Let X be an infinite dimensional locally convex space such that in equation the span of any compact metrizable subset has dimension <c. Then there is no strongly continuous supercyclic operator semigroup equationon X.

Proof: Assume that there exists a supercyclic strongly continuous operator semigroup equation on X. It is straightforward to verify that equation is a strongly continuous semigroup on equation . Pick three linearly independent vectors f1, f2 and f3 in X′. By Lemma 8.6, equation is finite dimensional for equation . Clearly each Ej equation -invariant for any equation Then equation is finite dimensional and equation -invariant for any equation Since fj∈ E for equation. Since E is equation invariant, we see that its annihilator equation if Tt -invariant for each equation. Thus we can consider the quotient operators equation Clearly equation is a strongly continuous operator semigroup on X/F. Moreover, equation is supercyclic since equation is. Now since equation is finite dimensional and has dimension ≥3. By Lemma 8.3, there are no strongly continuous supercyclic operator semigroups on X/F. This contradiction completes the proof.

Proof of Theorem 8.1: If X has no compact metrizable subsets whose linear span has dimension c, Lemma 8.6 implies that the linear span of any orbit equation is finite dimensional. It follows that equation is not supercyclic. It remains to consider the case when equationhas no compact metrizable subsets whose linear span has dimension c and apply Proposition 8.7.

The space φ

Recall that φ is a linear space of countable algebraic dimension carrying the strongest locally convex topology. In this section we mention certain properties of φ, mainly those which are related to continuous linear operators. It is well known [17] that φ is complete and all linear subspaces of φ are closed. Moreover, infinite dimensional subspaces of φ are isomorphic to φ. It is also well-known that for any topology θ on φ such that (φ,θ) is a topological vector space, θ is weaker than the original topology of φ. The latter observation immediately implies the following lemma.

Lemma 9.1: For any topological vector space X and any linear map equation T is continuous.

Lemma 9.2: Let X be a topological vector space and equation be a surjective continuous linear operator. Then X is isomorphic toequation

Proof. Since T is linear and surjective, there exists a linear map equation such that TS=I. By Lemma 9.1, S is continuous. Consider the linear maps

equation

It is easy to see that A and B are continuous and that AB=I and BA=I. Hence B is a required isomorphism.

Corollary 9.3: Let X be a topological vector space. Then the following conditions are equivalent. pt

(9.3.1) X is isomorphic to a space of the shape Y × φ,, where Y is a topological vector space ;

(9.3.2) X has a quotient isomorphic to φ;

(9.3.3) there is equation such that T(X) is infinite dimensional.

Proof: The implications (9.3.1)⇒(9.3.2)⇒(9.3.3) are trivial. Assume that (9.3.3) is satisfied. That is, there is equation with infinite dimensional T(X). Since any infinite dimensional linear subspace of φ is isomorphic to φ, we see that T(X) is isomorphic to φ. Hence there is a surjective equation By Lemma 9.2 X is isomorphic to Y × φ, where Y = ker S . Hence (9.3.3) implies (9.3.1), which completes the proof.

Cyclic operators on φ

Clearly φ is isomorphic to the space p of all polynomials over k endowed with the strongest locally convex topology. The shift operator on φ is obviously similar to the operator

equation(9.1)

For each equation we denote

equation (9.2)

 

Clearly pn is an n-dimensional subspace of p.

Lemma 9.4: An operator equation is cyclic if and only if T is similar to the operator M.

Proof: Clearly 1 is a cyclic vector for M. Hence any operator similar to M is cyclic. Now let equation be cyclic and equation be a cyclic vector for T. Then the vectorsequation are linearly independent. Indeed, otherwise their span is finite dimensional, which contradicts cyclicity equation is an algebraic basis of φ. It is easy to see then that the linear map equation is invertible and equation . By Lemma 9.1, J and J-1 are continuous. Hence T is similar to M.

We need a multicyclic version of the above lemma.

Lemma 9.5: Let equation . Then the following conditions are equivalent.

(9.5.1) T is multicyclic;

(9.5.1) there exists k∈equation and a linear subspace Y of φ of finite codimension such that equation and the restriction equation is similar to Mk, where M is the operator defined in (9.1).

Proof: First, assume that (9.5.2) is satisfied. Pick a finite dimensional subspace Z of such that equation. Since equation is similar to Mk,, we can pick an invertible linear operator equationsuch that equation for any equation. Let equation, where pk is defined in (9.8). Clearly L is a finite dimensional subspace of φ. From the equality equation it easily follows that for any equation

equation

Hence the linear span of the union of equation for equation contains equation.Thus T is m -cyclic with m=dimL. The implication (9.5.2)⇒(9.5.1) has been verified.

Assume now that T is n -cyclic for some equation. Then there is an n -dimensional subspace L of φ such that

equation (9.3)

(again, we use the fact that any linear subspace of φ is closed and therefore a dense subspace of φ must coincide with φ). We will use the concept of T -independence, introduced in Section 7. Since T -independence implies linear independence, any T -independent subset of L has at most n elements. Let k be the maximum of cardinalities of T -independent subsets of L and A be a T -independent subset of cardinality k. Since A is linearly independent, we can pick a subset equation of cardinality n-k such that equation is a basis in L. From the definition of k it follows that for any equationis not T independent and therefore using T -independence of A, we can find polynomials pb and pb,a for a∈ A such that

equation (9.4)

Now let m=0 if equation and equationotherwise. Consider the spaces

equation

Then Z is finite dimensional and equation. Obviously,

equation

Let equation and equation. Since equation and equation , we can find polynomials q,r such that equation and equation . Then

equation. Sinceequation . According to (9.4), equation. Since Y is invariant for equation. Thus

equation(9.6)

Since equation is a basis of L, from (9.5) and (9.6) it follows that equation for each equation.According to (9.3), equation . Since Z is finite dimensional, we see that Y has finite codimension in φ. In particular, Y is non-trivial and thereforeequation . That is, 1 ≤ k ≤ n and equation. Now consider the linear operator equation, which sends the monomial equation , where equation and equation are uniquely defined by the equationequation. By definition of Y, J is onto. From T -independence of A it follows that J is also one-to-one. By definition of J, we have equation. . Hence equation for any equation. That is, Mk and equation are similar.

Corollary 9.6: Let T be a multicyclic operator on φ. Then T is not onto.

Proof : According to Lemma 9.5, we can decompose φ into a direct sum equation , where Z has finite dimension equation , equationand equation is similar to Mk for some equation. Since equation is similar to Mk, equation has codimension equation in Y. Hence equation . On the other hand, equation.Thus equation has positive codimension in φ. Hence Tm+1 is not onto and so is T.

Proof of Theorem 1.12

In this section X is a topological vector space, which has no quotient isomorphic to φ. We have to show that there are no cyclic operators with dense range on X×φ. Assume the contrary and let equation be a cyclic operator with dense range. Consider the matrix representation of T:

equation

With T acting according to the formula equation Since T is cyclic, we can pick a vector equation such that equation is dense in X×φ. Since T has dense range, then Tm has dense range for any equation. Thus equationis dense in X×φ for each equation. Let equation for equation and uk =φ. Since equation is dense in X×φ, we see that equation is dense in φ for any equation. Hence Fm= φ for any equation. Since X has no quotients isomorphic to φ, Lemma 9.3 implies that equation is a finite dimensional subspace of φ. Then the space

equation

is also finite dimensional. Clearly equation and equation for any equation.It follows that each uk belongs to the space spanned by the union of equation for equation. Since L is finite dimensional, D is multicyclic. By Lemma 9.5, we can decompose φ into a direct sum equation, where Z is finite dimensional, equation and equation is similar to Mn for some equation. That is, there exists an invertible equation such that equation. Let also equation be the linear projection onto Y along Z. We consider two cases.

Case 1: The sequence equation is bounded from above. In this case equation is finite dimensional. Since J is invertible, equation is finite dimensional. Since P has finite dimensional kernel, equation is finite dimensional. We have arrived to a contradiction with the equality F0=φ.

Case 2: The sequence equation is unbounded from above. Since equation is finite dimensional,

equation

We shall show that equation whenever equation . Indeed, let equation be such that equation. By definition of P, equationand equation. As we know, equation . Hence equation. Since equation belong to, Y we have equation. Thus there is equation such that equation. Hence equation

Since equation, we have equation(the degree of the sum of two polynomials of different degrees equals to the maximum of the degrees). Since the sequence equation is unbounded from above, there is equationsuch that equationand according to the just proven statement, we will have equation for j ≥ k. Since any family of polynomials with pairwise different degrees is linearly independent, we see that the vectors JPuj for j ≥ k are linearly independent. Since JP is a linear operator, the vectors uj for j ≥ k are linearly independent in φ. Hence the sequence of spaces Fj for j ≥ k is strictly decreasing. On the other hand, we know that Fj=φ for eachequation. This contradiction completes the proof.

Hypercyclicity of operators on direct sums

We shall prove the following lemma, which is a key for the proof of Theorem 1.13.

Lemma 10.1: Let equationbe a sequence of infinite dimensional locally convex spaces such that

(10.1.1) there exists a dense linear subspace Y of X0, carrying a topology, stronger than the one inherited from X0 and turning Y into a separable metrizable topological vector space;

(10.1.2) there exists equationwith dense range;

(10.1.3) for each equation, there exists equationwith dense range.

Then there is a hypercyclic operator S on equation

It is worth noting that condition (10.1.1) implies that X0 is separable, condition (10.1.2) implies that X1 is separable and condition (10.1.3) implies that Xn for equation are all separable. Thus X is separable. We need the following auxiliary lemma.

Lemma 10.2: Let X and Y be topological vector spaces such that there exists equationwith dense range. Then for any closed hyperplane H of X, there exists S ∈ L(H,Y) with dense range.

Proof: Let equation be the restriction of T to H. We can write equation, where equationand equation. If T0 has dense range, thenequationis a continuous linear operator from H to Y with dense range. It remains to consider the case when the range of T0 is not dense. Since the range of T is dense and T0is a restriction of T to a closed hyperplane, the codimension of the closure of T0(H) in equation does not exceed 1. Thus the codimension of the closure of T0(H) in equation is exactly 1 and there is a non-zero equation such that T0(H) is a dense subspace of equation. If equation, then again we can take equation. Ifequation, we can pick y ∈ Y such that equation. Take equation. It is straightforward to verify that S has dense range.

Proof of Lemma 10.1: Let equation be a base of topology of Y and

equation

Clearly X is the union of the increasing sequence of subspaces Zn and each Zn is naturally isomorphic to equation. We shall construct inductively a sequence of operators equation and vectors equationsatisfying the following conditions for any equation:

(a1) equation

(a2) equation

(a3) equation

(a4) equation

(a5) equation

By (10.1.2) there exists equationwith dense range. Since Y is dense in X0=Z0, S0 has dense range and Z0 is nowhere dense in Z1, we can pick equation such that equation The basis of induction has been constructed. Assume now that equationand equationsatisfying (a1–a5) for k < n are already constructed. According to (a3) for k=n-1, we have equation. That is, the nth component wn of w is non-zero. Since Xn is locally convex, we can pick a closed hyperplane H in Xn such that equation.Let P be the linear projection on Zn onto H along equation. From (10.1.3) and Lemma 10.2 it follows that there is equationwith dense range. According to (a3) for k=n-1, the operator equation from Z0 to Zn has dense range. Hence the operator equation from Z0 to Xn+1 has dense range. Since Y is dense in Z0, we can pick equation such that equation. Now we define the operator equation . It is easy to see that equation. We set

equation

The operator Sn is continuous since Sn-1 and R are continuous. Clearly (a1) and (a5) for k=n are satisfied. Next, equation. Since equationis dense in Zn (condition (a2) for k=n-1) and equation is dense in Xn+1, we see that equation is dense inequation,which gives us (a2) for k=n. From the last display and the relation equation it follows that (a3) is satisfied for k=n. Finally, since equationfrom the definition of w we see that (a4) for k=n is also satisfied. Thus the inductive construction of Sk and Yk is complete.

Condition (a2) ensures that there is a unique operator S∈ L(X) such that equation for anyequation. From (a4) it now follows that

equation

According to the above display, the set equation is contained in the orbit equation. By (a5) A is dense in Y. Since Y is dense in X0 and carries a stronger topology, A is dense in X0=Z0. By (a2) equationis dense in Zm for each equation.Since equation , we have that equation and therefore equation is dense in Zm for each equation. Hence equationis dense in X. That is, Y is a hypercyclic vector for S.

Remark 10.3: The orbit of the hypercyclic vector constructed in the proof of Lemma 10.1 is not just dense. It is sequentially dense. The latter property is strictly stronger than density already for countable direct sums of separable infinite dimensional Banach spaces.

Proof of Theorem 1.14

Let equationfor each equation and equation.We shall apply Lemma 10.1. Condition (10.1.1) is satisfied according Lemma 3.3. Indeed X1 admits a Banach s -disk with dense span. By Lemma 3.8, spaces X0× X1 and equation belong to equation. From Lemma 3.6 it follows that conditions (10.1.2) and (10.1.3) are also satisfied. Thus by Lemma 10.1, there is hypercyclic T ∈ L(X).

Proof of Theorem 1.13

As we have already mentioned, s Fréchet space X belongs to equation if and only if X is infinite dimensional, separable and non-isomorphic to ω. Moreover, in any separable Fréchet space there is anequation -sequence with dense span.

Lemma 10.4: Let X and Y be separable infinite dimensional Fréchet spaces. Then the following conditions are equivalent

(10.4.1) there is no T ∈ L(X, Y) with dense range;

(10.4.2) X is isomorphic to ω and Y is not isomorphic to Ω.

Proof: If both X and Y are isomorphic to ω, then obviously there is a surjective T ∈ L(X,Y). If X is isomorphic to ω, Y is not and T ∈ L(X,Y), then Z=T(X) carries minimal locally convex topology [29] since ω does. It follows that Z is either finite dimensional or isomorphic to ω and therefore complete. Hence Z is closed in Y. It follows that equationsince Y is neither finite dimensional nor isomorphic to . Thus there is not T ∈ L(X,Y) with dense range. It remains to show that there is T∈ L(X,Y) with dense range if X is not isomorphic to ω. In this case the topology of X is not weak and it remains to apply Lemma 3.6.

Lemma 10.5: Let X be the countable locally convex direct sum of a sequence of separable Fréchet spaces infinitely many of which are infinite dimensional. Then one of the following two possibilities occurs:

(10.5.1) X is isomorphic to equation, where Y is a separable infinite dimensional Fréchet space and Z is the locally convex direct sum of an infinite countable number of copies of ω;

(10.5.2) X is isomorphic to equation , where each Yn is a separable infinite dimensional Fréchet space non-isomorphic to ω.

Proof: Separating the finite dimensional spaces, spaces isomorphic to ω and infinite dimensional spaces non-isomorphic to ω, we see that

equation

where the sets A, B and C are pairwise disjoint, Xα is isomorphic to ω for each equation is a separable infinite dimensional Fréchet space nonisomorphic to ω for any equation, Xγ is finite dimensional for each equationis infinite and countable and C is either finite or countable.

If B and C are finite, then A is infinite. Pick equation. Then

equation

Clearly Y is a separable infinite dimensional Fréchet space. Since each Xα is isomorphic to ω, we fall into the case (10.5.1). If B is finite and C is infinite, then A is infinite and both A and C can be enumerated by elements of equation.Then

equation

Again Y is a separable infinite dimensional Fréchet space. Since each equation is isomorphic to ω, (10.5.1) is satisfied. If B is infinite and equation is infinite, then we enumerate both B and equation by the elements of equation+: equation.We arrive to

equation

Since each equation is a separable infinite dimensional Fréchet space non-isomorphic to ω, (10.5.2) is satisfied. Finally, if B is infinite and equation is finite, we fix equation and write

equation

Again Z and each Xβ are separable infinite dimensional Fréchet spaces non-isomorphic to ω and we fall into the case (10.5.2).

We are ready to prove Theorem 1.13. Let X be a countable infinite direct sum of separable Fréchet spaces. If all the spaces in the sum, except for finitely many, are finite dimensional, then X is isomorphic to Y×φ, where Y is a Fréchet space. According to Theorem 1.12, X admits no cyclic operator with dense range. In particular, there is no supercyclic operator on X. If there are infinitely many infinite dimensional spaces in the sum defining X, then according to Lemma 10.5, we see that X is isomorphic to

equation

where Yn are all separable infinite dimensional Fréchet spaces and either all Yn are non-isomorphic to ω or all Yn for equation are isomorphic to . In any case from Lemma 10.4 it follows that there exists equationwith dense range and for each equation, there exists equationwith dense range. By Lemma 10.1, there is a hypercyclic operator on X. The proof of Theorem 1.13 is complete.

Hypercyclic operators on countable unions of spaces

The following lemma is a main tool in the proof of Theorem 1.11.

Lemma 11.1: Let a locally convex space be the union of an increasing sequence equation of its closed linear subspaces. Assume also that for any equation there is an equation-sequence with dense span in Xn and the topology of equation is not weak, where equation . Then there exists a linear map equation such that for any equation,equation and the orbit equation is dense in X.

Note that we do not claim continuity of the above operator S on X. Although if, for instance, X is the inductive limit of the sequence {Xn}, then continuity of S will immediately follow from the continuity of the restrictions equation.

Proof of Lemma 11.1: For each equation , let equation be an equation-sequence with dense span in Xn. For any equation, we apply Lemma 3.1 with the triple of spaces equation being (X,Xn,Xn−1) to obtain a sequence equation in equation such that equation is uniformly equicontinuous, each fn,k vanishes on Xn-1 and equation . According to Lemma 3.3, there exists a Banach disk K in X such that Xk is a dense subspace of X1 and the Banach space Xk is separable. Let equation be a base of topology of Xk. We shall construct inductively a sequence of operators equation and vectors equationsatisfying the following conditions for any equation:

(p1) equation

(p2)equation

(p3)equation

(p4)equation

(p5)equation

Consider the linear map equation defined by the formula

equation

Since equationis an equation-sequence in X2 and equationis uniformly equicontinuous, the above display defines a continuous linear operator from X to X2. Since equationcontains equation. Henceequationis dense in X2. Since Xk is dense in X1, S1 has dense range and equation is nowhere dense in X2, we can pick equation such that equation. The basis of induction has been constructed. Assume now that equationand equation, satisfying (p1–p5) for equation, are already constructed. According to (p3) for equation, we have equation where equation. Since equation is a closed hyperplane in Xn, we have equation. Let also equation Then H0 is a closed hyperplane of H. By (p2) for equation , the operator equation from X1 to Xn has dense range. Since Xk is dense in X1, we can pick equation such that equation. Thus equation. From definition of H0 it now follows that the matrix equationis invertible. The latter property allows us for any two vectors equation to find equation satisfying equation and equation . Pick any vector equation such that equation and let

equation

The above series converge since equation is an equation-sequence and equationis uniformly equicontinuous. Applying the above property to the pair equation , we find equation such that

equation

Consider now the linear map equation defined by the formula

equation

The above display defines a continuous linear operator since

equation is an equation-sequence and equation is uniformly equicontinuous. From the last three displays it follows that equation and equation. From definition of w and u and the relation equationit follows that (p3) and (p4) for k=n are satisfied. Clearly (p5) for k=n is also satisfied. Since each equationvanishes on Xn-1, we have from the last display that equation for any equation. Hence (p1) for k=n is satisfied. It remains to verify (p2) for k=n. Let U be a non-empty open subset of Xn+1. Since equation is dense in Xn+1, we can findequationand a convex balanced neighborhood W of zero in Xn+1 such that equation. Since equation, for each equation, we can pick equation such that equation and equation. It follows that equation By (p2) for equation is dense in Xn. Since equation, we can find equation such that equationBy the already proven property (p1) for k =n , equation.Hence equation. Using the equality equation, we get equation. Hence any non-empty open subset of Xn+1, contains elements of equation , which proves (p2) for k=n. Thus the inductive construction of Sk and yk is complete.

Condition (p2) ensures that there is a unique linear map equationsuch that equation for any equation. From (p4) it now follows that equation for each equation. Thus the set equation is contained in the orbit equation . By (p5) A is dense in Xk and therefore is dense in X1. By (p2) equation is dense in Xm+1for each equation. Since equation we have that equation and therefore equation is dense in Xm for each m ∈equation . Hence equationis dense in X.

Before proving Theorem 1.11, we need to make the following two elementary observations.

Lemma 11.2: Let X be an LB-space and Y be a closed linear subspace of X. Then either X /Y is finite dimensional or the topology of X /Y is not weak.

Proof: Since X is an LB-space, it is the inductive limit of a sequence equation of Banach spaces. If X /Y is infinite dimensional, we can find a linearly independent sequence equationin X′ such that each fn vanishes on Y. Next, we find a sequence equationof positive numbers converging to zero fast enough to ensure that equationas equationfor each equation.It follows that the sequence equation is pointwise convergent to zero on X. Since any LB-space is barrelled [17,29], the set equationis uniformly equicontinuous. Hence equationis a continuous seminorm on X. Since each fn vanishes on Y, equation. Then equationis a continuous seminorm on X /Y. Since fn are linearly independentequationhas infinite codimension in X and therefore equation has infinite codimension in X /Y. Hence the topology of X /Y is not weak.

Lemma 11.3: Let X be an inductive limit of a sequence equationof Banach spaces such that X0 is dense in X. Then X has no quotients isomorphic to φ.

Proof: Assume that X has a quotient isomorphic to φ. By Lemma 9.2 then X is isomorphic to Y×φ for some closed linear subspace Y of X. Let equationbe the natural embedding. Since X0is dense in X, J has dense range. Hence equation′ is injective. Since X is isomorphic to Y×φ, we have that equation′ is isomorphic to equation(ω is naturally isomorphic to equation). Hence, there exists an injective continuous linear operator from ω to the Banach space equation (with the topology equation ). That is impossible, since any injective continuous linear operator from ω to a locally convex space is an isomorphism onto image and ω is non-normable.

Proof of Theorem 1.11

Throughout this section X is the inductive limit of a sequence equation of separable Banach spaces. Let also equation be the closure of Xn in X. First, we shall prove the implicationequation. Assume that (1.11.4) is satisfied. Then we can pick a strictly increasing sequence equation of non-negative integers such that equation for each equation, Hence, for any equation, we can pick a non-trivial finite dimensional subspace Yk of equation such that equation. Thus the vector space X can be written as an algebraic direct sum

equation        (11.1)

 

Apart from the original topology τ on X, we can consider the topology θ, turning the sum (11.1) into a locally convex direct sum. Obviously equation . On the other hand, if W is a balanced convex θ-neighborhood of 0 in X, then equation is a τ-neighborhood of zero in equation for any equation. Indeed, it follows from the fact that equation where equation and Zk is finite dimensional. Since the topology of each is stronger than the one inherited from X, we see that equation is a neighborhood of zero in equation for each equation. Since X is the inductive limit of the sequence equation W is a τ -neighborhood of zero in X. Hence equation . Thus equation τ and therefore X is isomorphic to equation where Y is the locally convex direct sum of Yk for equation. Since Yk are finite dimensional, Y is isomorphic to φ. Since equation is the inductive limit of the sequence equation of separable Banach spaces (with the topology inherited from equation), the first one equation of which is dense, we see that (1.11.3) is satisfied. The implication equation is verified.

Assume now that (1.11.3) is satisfied. By Lemma 11.3, Y has no quotients isomorphic to φ. Theorem 1.12 implies now that there are no cyclic operators with dense range on X, which proves the implication equation. The implication equation is obvious since any hypercyclic operator is cyclic and has dense range. It remains to show that (1.11.1) implies (1.11.4). Assume the contrary. That is, (1.11.1) is satisfied and (1.11.4) fails. The latter implies that either there is n ∈equation+ such that equation is dense in X or there is a strictly increasing sequence equation of non-negative integers such that equation is infinite dimensional for each equation. In the first case, it is easy to see that X ∈ M and therefore there is a hypercyclic operator on X by Corollary 1.4. We have obtained a contradiction with (1.11.1). It remains to consider the case when there exists a strictly increasing sequence equation of non-negative integers such that equation is infinite dimensional for each equation. By Lemma 11.2, the topology of each equation is not weak. Let equation. Since equation is a separable Banach space, there is an equation1-sequence equation in equation with dense span. Since the topology on equation inherited from X is weaker than the Banach space topology of equation, equation is an equation1-sequence with dense span in equation . By Lemma 11.1, there exists a linear map equation and equation such that for any equation the restriction of S to equation is continuous and the orbit equation is dense in X. Since the topology of equation is stronger than the one inherited from X, we have that each restriction of S to equation is a continuous linear operator from equation to X. Since X is the inductive limit of the sequence equation, equation is continuous. Hence S is a hypercyclic continuous linear operator on X. The existence of such an operator contradicts (1.11.1). The proof of the implication equation and that of Theorem 1.11 is now complete.

Remarks on mixing versus hereditarily hypercyclic

We start with the following remark. As we have already mentioned, φ supports no supercyclic operator [11], which follows also from Theorem 1.12. On the other hand, φ supports a transitive operator [22]. The latter statement can be easily strengthened with the help of Corollary 5.1. Namely, take the backward shift T on φ. That is Te0=0 and equation for n ≥ 1, where equation is the standard basis in φ. Clearly T is a generalized backward shift and therefore T is an extended backward shift. By Corollary 5.1, I+T is mixing. Thus we have the following proposition.

Proposition 12.1 φ supports a mixing operator and supports no supercyclic operators.

On the other hand, a topological vector space of countable algebraic dimension can support a hypercyclic operator, as observed by several authors, [22], for instance. The following proposition formalizes and extends this observation.

Proposition 12.2 Let X be a normed space of countable algebraic dimension. Then there exists a hypercyclic mixing operator T∈ L(X).

Proof. Let equation be the completion of X. Then equation is a separable infinite dimensional Banach space. By Corollary 1.4, there is a hereditarily hypercyclic operator on equation. equation be a hypercyclic vector for S and E be the linear span of the orbit of x: equation. Grivaux [39] demonstrated that for any two countably dimensional dense linear subspaces E and F of a separable infinite dimensional Banach space Y, there is an isomorphism J :Y →Y such that J(E)= F . Hence there is an isomorphism equation such that J(X)= E . Let now equation. Since J(X)= E and E is S -invariant, X is T0-invariant. Thus the restriction T of T0 to X is a continuous linear operator on X. Moreover, since the S -orbit of s is dense in equation the T0-orbit of J-1x is dense in equation. Since equation , the latter orbit is exactly the T -orbit of J-1x and therefore J-1x is hypercyclic for T. Hence T is hypercyclic. Next, T0 is mixing since it is similar to the mixing operator S. Hence T is mixing as a restriction of a mixing operator to a dense subspace.

By Proposition 1.1, if X is a Baire separable and metrizable topological vector space, then any mixing T∈ L(X) is hereditarily hypercyclic. From the above proposition it follows that there are mixing operators on countably dimensional normed spaces. The next theorem however implies that there are no hereditarily hypercyclic operators on countably dimensional topological vector spaces, emphasizing the necessity of the Baire condition in Proposition 1.1.

Theorem 12.3 Let X be a topological vector space such that there exists a hereditarily universal family equation. Then equation. Proof. Since the topology of any topological vector space can be defined by a family of quasinorms [17], we can pick a nonzero continuous quasinorm P on X. That is, equation is nonzero, continuous, equation for any x,y ∈X, equationequation if x ∈ X , z ∈ K, | z |=1 and (X,τ p) is a (not necessarily Hausdorff) topological vector space, where τp is the topology defined by the pseudometric d(x, y) = p(x − y) . The latter property implies that equation for any t ∈K and any sequence equation in X such that equation Let κ be the first uncountable ordinal (commonly denoted ω1). We shall construct inductively sequences equation and equation of vectors in X and subsets of Z+ respectively such that for anyα < κ, pt

(s1) Aα is infinite and xα is a universal vector for the family equation;

(s2) equation as n→∞ , n ∈ Aα for any β < α;

(s3) Aα \ Aβ is finite for any β < α.

For the basis of induction we take A0=equation+ and x0 being a universal vector for the family {Tn : n ∈equation+}. It remains to describe the induction step. Assume that γ < κ and xα, Aα satisfying (s1–s3) for α < γ are already constructed. We have to construct xγ and Aγ satisfying (s1–s3) for α=γ.

Case 1: γ has the immediate predecessor. That is γ =ρ +1 for some ordinal ρ < κ. Since xp is universal for : equation , we can pick an infinite subset Aγ ⊂ Aρ such that equation as n→∞ , n∈Aγ . Since Aγis contained in Ap, from (s3) for α ≤ ρ it follows that Aγ \Aβ is finite for any β < γ. Hence (s3) for α=γ is satisfied. Now from (s3) for α=γ. and (s2) for α < γ. it follows that (s2) is satisfied for α=γ. Next, since {Tn : n ∈equation+} is hereditarily universal, we can pick xγ ∈ X universal for {Tn : n ∈ Aγ}. Hence (s1) for α=γ. is also satisfied.

Case 2: γ is a limit ordinal. Since γ is a countable ordinal, we can pick a strictly increasing sequence equationof ordinals such that equation Now pick consecutively n0 from equation , n1 > n0 from equation , n2 > n1 from equation etc. The choice is possible since by (s3) for α < γ, each equation is infinite. Now let equation. Since equation, equation Aα is finite for each j ∈equation+. Now if β < γ, we can pick j ∈equation+ such that β < αj < γ. Then equation is finite by (s3) with α=j. Moreover, since Aγ is contained in equation up to a finite set, from (s2) with α=αj it follows that p(T x )→0 as n→∞ , n∈Aγ . Hence (s2) and (s3) for α=γ are satisfied. Finally, since {Tn : n ∈equation+} is hereditarily universal, we can pick xγ ∈ X universal for {Tn : n ∈Aγ}. Hence (s1) for α=γ is also satisfied. This concludes the construction of {xa}α<κ and {Aa}α<κ satisfying (s1–s3).

In order to prove that equation , it suffices to show that vectors {xa}α<κ are linearly independent. Assume the contrary. Then there are n ∈N, equation and ordinals α1<…<αn<κ such that equation. By (s2) with α=αn, we see that equation as k →∞ , equation for 1 ≤ j < n. Denoting for equation 1 ≤ j < n and using linearity of Tk, we obtain equation for any k ∈equation+. Since p is a quasinorm, we have

equation

The above display contradicts universality of equation, equation which is (s1) with α= αn. This contradiction completes the proof.

Corollary 12.4 A topological vector space of countable algebraic dimension supports no hereditarily hypercyclic operators.

It is worth noting that there are infinite dimensional separable normed spaces, which support no supercyclic or transitive operators. We call a continuous linear operator T on a topological vector space X simple if T has shape T = zI + S , where z ∈equation and S has finite rank. Observe that a simple operator on an infinite dimensional topological vector space is never transitive or supercyclic. Indeed, let T be a simple operator on an infinite dimensional topological vector space and λ ∈K, S∈ L(X) be such that T =λ I + S and S has finite rank. Then L=S(X) is finite dimensional. Since X/Y is infinite dimensional, we can pick non-empty open subsets U0 and V0 of X/L such that equation does not intersect equation . Let equation and equation. Clearly U and V are non-empty open subsets of X. Using the equalities T =λ I + S and S(X)=L, it is easy to see that equation for any equation. Hence T is non-transitive. Moreover since U and V are stable under multiplication by non-zero scalars, the projective orbit equation of any x ∈ U does not meet V. Hence U contains no supercylic vectors for T. Since the set of supercyclic vectors of any continuous linear operator is either dense or empty, T is non-supercyclic.

We say that a topological vector space X is simple if it is infinite dimensional and any T∈ L(X) is simple. Thus simple topological vector spaces support no supercyclic or transitive operators. Various examples of simple separable infinite dimensional normed spaces can be found in the literature [38,40-44]. Moreover, according to Valdivia [40], in any separable infinite dimensional Fréche spaced there is a dense simple hyperplane. All the examples of this type existing in the literature with one exception [44] are constructed with the help of the axiom of choice and the spaces produced are not Borel measurable in their completions. In [44] there is a constructive example of a simple separable infinite dimensional pre-Hilbert space H which is a countable union of compact sets.

Finally recall that an infinite dimensional topological vector space X is called rigid if L(X) consists only of the operators of the form I for λ ∈equation. Of course, a rigid space can not be locally convex. Clearly there are no transitive or cyclic continuous linear operators on a rigid topological vector space. Since there exist rigid separable equationspaces [45], we see that there are separable infinite dimensional equationspaces on which there are no cyclic or transitive operators.

Concluding remarks and open problems

We start by observing that the following questions remain open.

Problem 13.1 Is there a hereditarily hypercyclic operator on a countable direct sum of separable infinite dimensional Banach spaces?

Problem 13.2 Is there a hypercyclic strongly continuous operator semigroup on a countable direct sum of separable infinite dimensional Banach spaces?

The most of the above results rely upon the underlying space being locally convex or at least having plenty of continuous linear functionals and for a good reason. As mentioned in the previous section, there are separable infinite dimensional equationspaces on which there are no cyclic or transitive operators. On the other hand, the absence of non-zero continuous linear functionals on a topological vector space does not guarantee the absence of hypercyclic operators on it. It is well-known [45] that the spaces Lp[0,1] for 0 ≤ p <1 are separable equationspaces having no non-zero continuous linear functionals. Ansari [9] raises a question whether these spaces support hypercyclic operators. Theorem 1.18 provides an easy answer to this question. Namely, consider the operator T ∈L(Lp[0,1]) , Tf (x)= f (x / 2) . It is straightforward to see that T is onto and has dense generalized kernel. Thus I+T is hereditarily hypercyclic according to Corollary 5.2.

It is obvious that an extended backward shift has dense range and dense generalized kernel. Unfortunately, the converse is not true in general. This leads naturally to the following question.

Problem 13.3 Let T be a continuous linear operator on a separable Banach space, which has dense range and dense generalized kernel. Is it true that I+T is mixing or at least hypercyclic?

From Corollary 5.1 and Corollary 2.14 it follows that if T is an extended backward shift on a separable infinite dimensional Banach space X, then both I+T and eT are hereditarily hypercyclic. This reminds of the following question raised by Bermúdez, Bonilla, Conejero and Peris in reference [15].

Question B2CP. Let X be a complex Banach space and T∈ L(X) be such that its spectrum σ (T) is connected and contains 0. Does hypercyclicity of I+T imply hypercyclicity of eT? Does hypercyclicity of eT imply hypercyclicity of I+T?

We show that the answer to both parts of the above question is negative. Before doing this we would like to raise a similar question, which remains open. I+T

Problem 13.4 Let X be a Banach space and T∈ L(X) be quasinilpotent. Is hypercyclicity of I+T equivalent to hypercyclicity of eT?

First, we introduce some notation. Let equation be the Hardy Hilbert space on the unit disk and equation be the space of bounded holomorphic functions equation. For any equation , the multiplication operator equation is a bounded linear operator on equation . It is also clear that equation. if equation is the adjoint of Mφ, then equation is the reflection of σ (Mφ ) with respect to the real axis. The following proposition is a direct consequence of a theorem by Godefroy and Shapiro [27][Theorem 4.9].

Proposition 13.5 Let equation. Then equation is hypercyclic if and only if

equation                (13.1)

 

The above Proposition calls for the following comment. A bounded linear operator T on a separable infinite dimensional Banach space X is said to satisfy the Kitai Criterion [46,47] if there exist dense subsets E and F of X and a map S : F →F such that TSy = y for any y∈F , Tnx→0 and equation as n→∞ for any x∈E and y∈F . As it is shown in [46,47], any operator satisfying the Kitai Criterion is hypercyclic. Moreover, any operator, satisfying the Kitai Criterion is hereditarily hypercyclic and therefore mixing [3]. Hypercyclicity in the proof of the above result in reference [27] is demonstrated via application of the Kitai Criterion. Thus the following slightly stronger statement holds.

Corollary 13.6 Let equation. Then equation is hypercyclic if (13.1) is satisfied and equationis non-hypercyclic if (13.1) is not satisfied.

Now we demonstrate that the answer to both parts of Question B2CP is negative. Consider the subset U of equation being the interior of the triangle with vertices -1, I and -i. In other words equation Next, let equation . The boundary of V consists of the interval [−1+i,1+i] and two circle arcs. It is clear that U and V are bounded, open, connected and simply connected. From the definition of the sets U and V it immediately follows that the open set 1+U ={1+ z : z∈U} intersects the unit circle. On the other hand, since U is contained in the left half-plane, we see that equation Similarly, we see that equation and the open set eV intersects the unit circle. According to the Riemann Theorem [48], there exist holomorphic homeomorphisms equation and equation . Obviously equation . Since equation equation and both equation and equation intersect the unit circle, Corollary 13.6 implies that equation and equation are hereditarily hypercyclic. Since equation, equation , equation is contained in equation, and equation does not intersect equation , Corollary 13.6 implies that equation and equation are non-hypercyclic. Finally, observe that equation is the closure of U and equation is the closure of -V and therefore the spectra of equation and equation are connected and contain 0. Taking into account that all separable infinite dimensional Hilbert spaces are isomorphic, we arrive to the following result, which answers negatively the Question B2CP.

Proposition 13.7 There exist bounded linear operators A, B on the complex Hilbert space equation2 such that σ(A) and σ(B) are connected and contain 0, I+A and eB are hereditarily hypercyclic, while eA and I+B are non-hypercyclic.

Finally, if the answer to Question 13.4 is affirmative, then the following interesting question naturally arises.

Problem 13.8 Let A be a quasinilpotent bounded linear operator on a complex Banach space X and f be an entire function on one variable such that equation . Is it true that hypercyclicity of f(A) is equivalent to hypercyclicity of I+A?

Spaces Ck(M) and their duals

Let (M,d) be a separable metric space and Ck(M) be the space of continuous functions equation with the compact-open topology (=the topology of uniform convergence on compact subsets of M). It is easy to see that Ck(M) is complete. Moreover, Ck(M) is metrizable if and only M is locally compact and Ck(M) carries weak topology if and only if M is discrete. On the other hand, there always is an equation1- sequence with dense span inCk(M). Indeed, let equation be a metrizable compactification of M. Since equation is a separable Banach space, there is an equation1- sequence equation with dense span in equation . Since equation is densely and continuously embedded intoCk(M), equation is an equation1- sequence with dense span in Ck(M). Thus Ck(M) ∈M if and only if M is non-discrete. Corollary 1.4 implies now the following proposition.

Proposition 13.9 If (M,d) is a separable non-discrete metric space, then there is a hereditarily hypercyclic operator on Ck(X).

The reason for inclusion of the above proposition is to demonstrate that Theorem 1.3 and Corollary 1.4 are applicable far beyond metrizable or LB-spaces. The spaces Ck(M) can have quite ugly structure indeed. For instance, take M being the set equation of rational numbers with the metric induced from equation, and you have got the space Ck(equation), which does not fall into any of the well-understood and studied classes of locally convex spaces. We would like to raise the following question.

Problem 13.10 Characterize separable metric spaces (M,d) such that Ck(M) supports a dual hypercyclic operator.

It is worth noting that if M is discrete, then either Ck(M) is finite dimensional or is isomorphic to ω and therefore does not support a dual hypercyclic operator (there are no hypercyclic operators on φ=ω). In general, equation (M)=(Ck (M))′ can be naturally identified with the space of finite equation -valued Borel σ -additive measures on M with compact support. This dual space is separable in the strong topology if and only if all compact subsets of M are finite or countable. If it is not the case, there is no point to look for dual hypercyclic operators on Ck(M). Thus the only spaces (M,d) for which Ck(M) has a chance to support a dual hypercyclic operator are non-discrete spaces with no uncountable compact subsets. The first natural candidate to consider is equation.

Note also that although Theorem 1.15 provides answers to the questions of Petersson, mentioned in the introduction, it does not characterize Fréchet spaces, supporting a dual hypercyclic operator.

Problem 13.11 Characterize Fréchet spaces X such that X supports a dual hypercyclic operator.

The most natural Fréchet space for which we do not know whether it supports a dual hypercyclic operator is the countable power of the Hilbert space equation2.

The Hypercyclicity Criterion

The following universality criterion is proved by Bés and Peris [2, Theorem 2.3 and Remark 2.6]. It is formulated in [2] in the case when X is an equationspace, but the proof works without any changes for Baire separable metrizable topological vector spaces.

Theorem BP Let equation be a sequence of pairwise commuting continuous linear operators with dense range on a Baire separable metrizable topological vector space X. Then the following conditions are equivalent:

(a) The family equation is universal;

(b)There exists an infinite subset A of equation+ such that the familyequationis hereditarily universal

(c)There exist a strictly increasing sequence {nk} of non-negative integers, dense subsets E and F of X and maps equation for k ∈ equation+ such that equation, equation and equation as k→∞ for any x ∈ E and y ∈ F.

We formulate now the so-called Hypercyclicity and Supercyclicity Criteria, which follow easily from the above theorem.

Theorem HC Let X be a Baire separable metrizable topological vector space and T∈ L(X). Then the following conditions are equivalent:

(a)T ⊕ T is hypercyclic;

(b)There exists an infinite subset A of equation+ such that the family equation is hereditarily universal;

(c)There exist a strictly increasing sequence {nk} of non-negative integers, dense subsets E and F of X and maps equation for k ∈ equation+ such that ,equation, equationand equation as k→∞ for any x ∈ E and y ∈ F.

Theorem SC. Let X be a Baire separable metrizable topological vector space and T∈ L(X). Then the following conditions are equivalent:

(a) T ⊕ T is supercyclic;

(b) There exists an infinite subset A of equation+ and a sequence equation of positive numbers such that the family equation is hereditarily universal;

(c) There exist a strictly increasing sequence {nk} of non-negative integers, dense subsets E and F of X, and a sequence equation of positive numbers and maps equation for k ∈ equation+ such that and equation, equation and equationas k→∞ for any x ∈ E and y ∈ F.

An operator satisfying the condition (c) of Theorem HC (respectively Theorem SC) is said to satisfy the Hypercyclicity (respectively, Supercyclicity) Criterion. The long standing question whether any hypercyclic operator T on a Banach space satisfies the Hypercyclicity Criterion, was recently solved negatively by Read and De La Rosa [49]. Their result was extended by Bayart and Matheron [50], who demonstrated that on any separable Banach space with an unconditional Schauder basis such that the forward shift operator associated with this basis is bounded, there is a hypercyclic operator T such that T ⊕ T is not hypercyclic. This leaves open the following question raised in reference [50].

Problem 13.12 Does there exist a separable infinite dimensional Banach space X such that any hypercyclic operator on X satisfies the Hypercyclicity Criterion?

It is observed in reference [50] that any T∈ L(ω) satisfies the Hypercyclicity Criterion. It also follows from Theorem 1.7 and Theorem HC. Thus the above question in the class of Fréchet spaces has an affirmative answer, which leads to the following problem.

Problem 13.13 Characterize separable infinite dimensional Fréchet spaces X on which any hypercyclic operator on X satisfies the Hypercyclicity Criterion?

It is worth noting that non-hypercyclicity of T ⊕ T in references [49,50] is ensured by the existence of a non-zero continuous bilinear form b : X × X →equation with respect to which T is symmetric: b(Tx, y) = b(x,Ty) for any x, y∈X . The following proposition formalizes the corresponding implication. Similar statements have been proved by many authors in various particular cases. The proof goes along the same lines as in any of them.

Let X and Y be topological vector spaces and b : X× X→Y be separately continuous and bilinear. We say that T∈ L(X) is b-symmetric if b(Tx, y) = b(x,Ty) for any x, y∈X . Recall also that b is called symmetric if b(x, y)=b(y, x) for any x, y∈X and b is called antisymmetric if b(x, y) = −b(y, x) for any x, y∈X .

Proposition 13.14 Let X and Y be a topological vector spaces, b : X× X→Y be separately continuous, non-zero and bilinear and T∈ L(X) be b -symmetric. Then T ⊕ T is non-cyclic. If additionally b is nonsymmetric, then T2 is non-cyclic.

Proof. Consider the left and right kernels of b:

equation(13.2)

 

Separate continuity of b implies that equation are closed linear subspaces of X. Since b is non-zero, we have X0 ≠ X and X1 ≠ X . From b -symmetry of T it follows that X0 and X1 are both T -invariant. Hence X0× X and X × X1 are T ⊕ T -invariant proper closed subspaces of X × X, they can not contain a cyclic vector for T ⊕ T. Assume that T ⊕ T has a cyclic vector (x, y)∈X × X . Then equation and equation Consider now a continuous linear operator Φ: X× X →Y defined by the formula

equation

Since equation and equation we have equation On the other hand, using b -symmetry of T, we have

equation

Thus the orbit of (x, y) with respect to T T lies in the proper closed linear subspace ker Φ of X, which contradicts cyclicity of (x, y) for TT.

Assume now that b is non-symmetric. Then equation is a non-zero separately continuous bilinear map from X × X to Y. Moreover, T is c-symmetric. Assume that x is a cyclic vector for T2. Then x can not lie in the right kernel of c, which is a proper closed T-invariant subspace of X. Hence the operator equation, equation is nonzero. On the other hand, for any n ∈equation,

equation

Hence the orbit of x with respect to T2 lies in the proper closed linear subspace equation of X, which contradicts cyclicity of x for T2

This looks like a proper place to reproduce the following question of Grivaux.

Problem 13.15 Let X be a Banach space and T∈ L(X) be such that T ⊕ T is cyclic. Does it follow that T2 is cyclic?

As a straightforward toy illustration of the above proposition one can consider the following fact. Let equation be a measure space, equation, 0 < p < ∞ and equation be the operator of multiplication by g : Tf = fg for equation. Then T⊕ T is non-cyclic. Indeed, consider the continuous bilinear map equation. Clearly b is non-zero and T is b-symmetric. By Proposition 13.14, TT is non-cyclic. The above mentioned result of Bayart and Matheron can now be formulated in the following way.

Theorem BM Let X be a separable infinite dimensional Banach space with an unconditional Schauder basis such that the forward equation shift operator associated with this basis is bounded. Then there exists a hypercyclic T∈ L(X)and a non-zero continuous bilinear form such that T is b -symmetric. In particular, T T is non-cyclic.

The form b in the above theorem must be symmetric. Indeed, otherwise, by Proposition 13.14, T2 is non-cyclic, which contradicts hypercyclicity of T according to the Ansari theorem [1] on hypercyclicity of powers of hypercyclic operators. An answer to the following question could help in better understanding of the phenomenon of hypercyclic operators not satisfying the Hypercyclicity Criterion.

Problem 13.16 Let T be a hypercyclic continuous linear operator on a Banach space X such that T T is non-hypercyclic. Does there exist a non-zero symmetric continuous bilinear form equation such that T T is b -symmetric?

It is worth noting that non-existence of such a form b is equivalent to the density of the range of the operator equation projective tensor product equation Another observation concerning Theorem BM is that operators constructed in reference [50] have huge spectrum. Namely, their spectrum contains a disk centered at 0 of radius >1. On the other hand, we know (see Theorems 1.19 and 1.21) that any separable infinite dimensional complex Banach space supports plenty of hypercyclic operators with the spectrum being the singleton {1}. This leads to the following question.

Problem 13.17 Let T be a hypercyclic continuous linear operator on a complex Banach space X such that σ(T)= {1}. Is T T hypercyclic?

It is worth noting that an affirmative answer to the above question would take care of Problem 13.7. Indeed, the spectrum of any hypercyclic operator on a hereditarily indecomposable complex Banach space [51] is a singleton {z} with z ∈equation

n-supercyclic operators

Recently Feldman [52] has introduced the notion of an n -supercyclic operator for n ∈equation A bounded linear operator T on a Banach space X is called n-supercyclic for n ∈equation if there exists an n -dimensional linear subspace L of X such that its orbit equation is dense in X. Such a space L is called an n - supercyclic subspace for T. Clearly,1- supercyclicity coincides with the usual supercyclicity. In reference [52], for any n ∈equation, n ≥ 2, a bounded linear operator T on equation is constructed, which is n -supercyclic and not (n-1)-supercyclic. The construction is based on the observation that if Tk for equation are bounded linear operators on Banach spaces Xk all satisfying the Supercyclicity Criterion with the same sequence {nk}, then the direct sum T1 ⊕…⊕ Tn is n -supercyclic. This observation leads to the natural question whether the direct sum of n supercyclic operators should be n -supercyclic. The following proposition provides a negative answer to this question.

Proposition 13.18 There exists a hypercyclic operator T∈ L(equation) such that T T is not n -supercyclic for any n ∈equation

The above proposition follows immediately from Theorem BM and the next proposition, which implies that for the operator T from Theorem BM, TT is not n -supercyclic for any n ∈equation.

Proposition 13.19 Let X be an infinite dimensional topological vector space, equation be non-zero, separately continuous and bilinear and T∈ L(X) be a b-symmetric operator with no non-trivial closed invariant subspaces of finite codimension. Then for any finite dimensional linear subspace L of X × X, the set equation is nowhere dense in X × X.

In order to prove Proposition 13.19, we need the following lemma.

Lemma 13.20 Let m ∈ equation, L and X be topological vector spaces, such that equation and equation be a continuous bilinear map such that for each non-zero n ∈ L, the linear map equation is surjective. Then the set equation is closed and nowhere dense in X.

Proof. First, observe that it is enough to prove the required statement in the case, when X is finite dimensional. Indeed, let e1,...,ek be a basis of L and equationThen X1 is a closed linear subspace such thatequation for any equation. Moreover, X1 has codimension at most km in X. Thus we can pick a finite dimensional subspace Y of X such thatequation. Since equation for any equation, we have that for each equation , the restriction of equation to Y is onto. It is also clear that equation , where equation and π is the projection in X onto Y along X1. Since π is continuous and open, A is closed and nowhere dense in X if and only if A0 is closed and nowhere dense in Y, which is finite dimensional.

Thus without loss of generality, we can assume that X is finite dimensional. Consider the unit sphere S in L with respect to some Hilbert space norm on L. For each u S, equation is onto and we can pick an m-dimensional subspace Zu of X such that the restriction of equation to Zu is invertible. Clearly, the set Vu of those vS for which the restriction of equation to Zu is invertible is open in S and contains u. Thus we can pick a neighborhood Wu of u such that equation is onto for each equation. Since uWu,the family {Wu : uS} is an open cover of the compact space S and therefore we can choose u1,…, urS such that equation Then equation

It suffices to show that each Aj is closed and nowhere dense. Let equationCloseness of Aj is rather easy. Indeed, let equationbe a sequence of elements of Aj converging to xX. Since xnAj, we canequationsuch that equation. Since equationis compact, we, passing to a subsequence, if necessary, can assume that equationSince B is continuous, we have equation.Thus equationand xAj . That is, Aj is closed. It remains to show that it is nowhere dense. Pick a linear subspace Yj of X such that equation. For each equationlet Tu be the restriction ofequationto equationand Su be the restriction of equationto Yj. Let equationand yYj. Then equationif and only if there exists equationsuch that equation. Since Tu is invertible, the latter is equivalent to equation. Thus equation,whereequationLet n=dim X. Then equation . Hence equation is a manifold of dimension equationwhere α =1 if equationand α =1 if equationIt is clear that F is smooth and therefor is Lipschitzian on any compact set. Since a Lipschitzian map does not increase the Hausdorff dimension, we see that equationis a countable union of compact sets of Hausdorff dimension at most equation.Since equation,any compact subset of X of Hausdorff dimension <αn is nowhere dense. Thus Ajis a Baire first category set. Since Aj is closed, it is nowhere dense.

Proof of Proposition 13.9. First, we consider the case of nondegenerate b. That is, we assume that both the left and the right kernels X0 and X1 of b defined by (13.2) are trivial. For each k ∈equation and equation , consider the linear functional equation defined by the formula

equation

First, we shall check that for any (x, y) ≠ (0,0) , the functionals Φk (x, y) for k ∈ equation are linearly independent. Assume the contrary. Then there exists a non-zero polynomial p such that b(p(T)x,v)=b(u, p(T)y) for any u,vX. Since the left-hand side of the last equality does not depend on u and the right-hand side does not depend on v, they both do not depend on both u and v. Hence b(p(T)x,v) = b(u, p(T)y) = 0 for any u,v X. Since T is b -symmetric, we have b(x, p(T)v)=b(p(T)u, y)= 0 for any u,vX. Hence equation, where equation,equation.Since T has no non-trivial closed invariant subspaces of finite codimension, T' has no non-trivial finite dimensional invariant subspaces. By Lemma 7.2, p(T' ) is injective. Hence equation.Since b is non-degenerate, we then have x=y=0, which contradicts with the assumption equation. Thus the functionals equationfor k ∈equation are linearly independent for each equation.

Let L be a finite dimensional linear subspace of X × X, equation and m ∈ equation, m > n. Consider the bilinear map

equation

Since for any non-zero equation,the functionals equation are linearly independent, we see that the linear map equation is onto. By Lemma 13.20, the set

equation

is closed and nowhere dense in X × X. Let now equationfor some p ∈equationThenequation for some equation Then equationfor any j equationsince T is b-symmetric. It follows that (u,v)∈A . That is, the nowhere dense set A contains the set equation. Thus the latter set is nowhere dense.

It remains to reduce the general case to the case of nondegenerate b. Since b is non-zero, at least one of the bilinear forms equation or equation is non-zero. Clearly T is symmetric with respect to both b0 and b1. Thus replacing b by either b0 or b1, if necessary, we can assume that b is symmetric or antisymmetric. Then left and right kernels X0 and X1 of b defined by (13.2) coincide. Since T is b -symmetric, X0 is T-invariant. Let equation be defined as equation and β be the bilinear form on X / X0 defined by the formula equation . Since X0 is the right and the left kernel of b and is T -invariant, the operator T0 and the form β are well-defined and β is non-degenerate. Moreover, it is easy to see that T0 is β-symmetric and has no non-trivial closed invariant subspaces of finite codimension. Assume now that L a finite dimensional linear subspace L of X × X, equation and equation. According to the first part of the proof, B is nowhere dense in X / X0. Since equation, where equationis the canonical map from X ontoX / X0, we see that A is nowhere dense in X.

The following question is raised by Bourdon, Feldman and Shapiro [53].

Problem 13.21 Let X be a complex Banach space, n ∈ equation and T ∈ L(X) be such that T is n-supercyclic and equation Is T cyclic?

It is worth noting that the only known examples of n -supercyclic operators T with equationare the mentioned direct sums of operators satisfying the Supercyclicity Criterion with the same sequence {nk}. Such direct sums are all cyclic [54].

Feldman in reference [52] has also introduced the concept of an ∞-supercyclic operator. A bounded linear operator T on a Banach space X is called ∞ -supercyclic if there exists a linear subspace L of X such that its orbit {Tn x: n ∈ equation, t ∈ L} is dense in X, the space Tn(L) is not dense in X for any n ∈ equationand L contains no non-zero invariant subspace of T. Gallardo and Motes-Rodriguez [55], answering a question of Salas, demonstrated that the Volterra operator

equation

is not supercyclic. In [56] it is shown that V is not n -supercyclic for any n ∈ equation.However, it turns out that V is ∞-supercyclic.

Proposition 13.22 The Volterra operator is ∞-supercyclic.

Proof. For any non-zero hL2[0, 1] we denote by Lh the orthocomplement of h: equation. It is straightforward to see that for any hL2[0, 1],

equation (13.3)

is the adjoint of V. Consider the space equation If h is non-zero element of ε, then according to the above display, equation . Thus by (13.3)

equation (13.4)

Consider now the following specific h ∈ ε : h(1) =0 and equationfor 0 ≤ x <1 . First, we shall show that Lh does not contain any non-zero invariant subspace of V. Assume the contrary. Then there exists non-zero fL2[0, 1], such that Vn fLh for each nequation. That is, equation.Hence h is not a cyclic vector for V*. On the other hand, it is well-known that gL2[0, 1] is non-cyclic for V* if and only if there is q ∈ (0,1) such that g vanishes on [q, 1]. Thus h vanishes on a neighborhood of 1, which is obviously not the case. Hence Lh does not contain any non-zero invariant subspace of V. According to (13.4), equation is a closed hyperplane in L2[0, 1] and therefore Vn(Lh) is not dense in L2[0, 1] for each nequation. Now, in order to show that V is ∞-supercyclic it suffices to verify that equation is dense in L2[0, 1]. By (13.4), equation. It is easy to see that for any fL2[0, 1] and non-zero gL2[0, 1], the distance from f to Lg is given by the formula equation.et q ∈ (0,1) and fL2[0, 1] be such that f vanishes on [q, 1]. Then

equation

On the other hand, analyticity of h on [0, q] and easy lower estimates if equation imply that

equation

From the last two displays it follows that equation as n → ∞ . Hence equationcontains the space of all functions vanishing on a neighborhood of 1. Since the latter space is dense in equation , A is dense in equation, which completes the proof.

equation- cyclicity and supercyclicity

 

Let T be a continuous linear operator on a separable complex topological vector space X. We say that T is equation - cyclic if there exists x X such that the linear span of the orbit {Tn x: n ∈ equation} in X considered as a linear space over equation is dense in X. Similarly, T is called equation- supercyclic if there is xX such that {tTn x: nequation, tequation} is dense in X and T is called equation+- supercyclic if there is xX such that {tTn x: nequation, t > 0}is dense in X. Clearly any equation+-supercyclic operator is equation-supercyclic and any equation-supercyclic operator is equation -cyclic. The following theorem by León- Saavedra and Müller is proved in reference [57]. It is proved in the case when X is a Banach space, but exactly the same proof works in general.

Theorem LM Let T be a continuous linear operator on a complex locally convex space X with equation . Then T is supercyclic if and only if T is equation+-supercyclic.

As we have shown, there are bilateral weighted shifts Tw on equation(equation) with the weight sequence w converging to zero arbitrarily fast and such that I+Tw and I+T'w are both hypercyclic. This happens because we allow w to behave irregularly while still satisfying the condition | wn |≤ a|n| for every nequation with a being any sequence of positive numbers. The following proposition shows that hypercyclicity of I+Tw is incompatible with the symmetry of the weight sequence. Recall that a continuous linear operator on a Banach space X is called weakly supercyclic if it is supercyclic on X with weak topology.

Proposition 13.23 Let equation be a weight such that equation for any n ∈ equation, and let p be a polynomial with real coefficients. Then the operator p(Tw) acting on complex equation(equation) is not equation-cyclic. In particular, by Theorem LM, p(Tw) is not weakly supercyclic.

Corollary 13.24 Let equation with equation for any n ∈ equation Then the operator I+Tw acting on equation(equation) is not weakly supercyclic.

Proof of Proposition 13.23. First, note that if equation satisfy equation for any n ∈ equation, then Tw and equationare isometrically similar with a diagonal unitary operator implementing the similarity. Thus we can, without loss of generality, assume that wn equation for each n ∈ equation. Then the operators Tw and S=p(Tw) have real matrix coefficients with respect to the canonical basis. Let H be the equation-subspace of equation(equation) consisting of the sequences with real entries. Then equation and equation. Let T0 be the restriction of Tw to H and S0 the restriction of S to H considered as equation-linear operators. Since T0 and S0 are similar to the restrictions of Tw and S to iH, we see that Tw and S, considered as equation -linear operators, are similar to T0T0 and S0S0. Now the symmetry of the weight sequence w implies that T0 is isometrically similar to T'0. Indeed,equation, where equation for nequation. Then equation is similar to equation Thus S considered as an equation-linear operator is similar to S0S'0. The last operator is non-cyclic. Indeed, if xy is a non-zero vector in HH, then the orbit of xy with respect to S0S'0 is orthogonal to the non-zero vector y ⊕ (-x) ∈ HH. Thus S is not equation -cyclic.

Since equation(equation) for 1 ≤ p ≤ 2 is contained in equation(equation) and carries a stronger topology than the one inherited from equation(equation), Proposition 13.23 remains true if we replace equation(equation) by equation(equation) with 1 ≤ p ≤ 2 . On the other hand, the unweighted shift on equation(equation) with 2< p < ∞ is weakly supercyclic [37] and therefore the statement of Proposition 13.23 becomes false if we replace equation(equation) by equation(equation) with p > 2. At this point it is interesting to remind that, according to Theorem B, norm hypercyclicity and supercyclicity of a bilateral weighted shift on equation(equation) do not depend on p. This leads naturally to the following question.

Problem 13.25 Characterize hypercyclicity and supercyclicity of the operators of the form I+T, where T is a bilateral weighted shift on equation(equation). In particular, does hypercyclicity or supercyclicity of these operators depend on the choice of P. 1 ≤ p < + ?

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